Kinematics
Kinematics
Kinematics
KINEMATICS
THEORY AND EXERCISE BOOKLET
CONTENTS
3. Graphs ............................................................................. 15 – 26
6. Exercise -I ........................................................................ 53 – 70
7. Exercise - II ...................................................................... 71 – 76
9. Exercise - IV ..................................................................... 87 – 89
394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564
1 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053 www.motioniitjee.com, email-hr.motioniitjee@gmail.com
Page # 2 KINEMATICS
KINEMATICS
does not mean that our study will be restricted to small corpuscles;
motion and the concept of the relative motion of one particle with
IIT-JEE Syllabus :
Relative Motion.
(C) Distance :
The length of the actual path traversed by the particle is termed as its distance.
Distance = length of path ACB.
* Its SI unit is metre and it is a scalar quantity.
* It can never decrease with time.
(D) Displacement :
The change in position vector of the particle for a given time interval is known as its displacement.
→
AB = r = r2 − r1
* Displacement is a vector quantity and its SI unit is metre.
* It can decrease with time.
For a moving particle in a given interval of time
* Displacement can be +ve, –ve or 0, but distance would be always +ve.
* Distance ≥ Magnitude of displacement.
* Distance is always equal to displacement only and only if particle is moving along a straight line
without any change in direction.
(E) Average speed and average velocity :
Average speed and average velocity are always defined for a time interval.
Total dis tan ce travelled ∆s
Average speed(vav ) = =
Time int erval ∆t
Displacement ∆r r −r
Average velocity (vav ) = = = 2 1
Time int erval ∆t t2 − t1
* Average speed is a scalar quantity, while average velocity is a vector quantity. Both have the same
SI units, i.e., m/s.
For a moving particle in a given interval of time
* Average speed can be a many valued function but average velocity would be always a single-
valued function.
* Average velocity can be positive, negative or 0 but average speed would be always positive.
Distance
As ∆t tends to zero, the ratio defining speed becomes
finite and equals to the first derivative of the distance.
The speed at the moment 't' is is called the instantaneous D ∆S
speed at time 't'. θ
On the distance - time plot, the speed is equal to the slope A C
∆t
of the tangent to the curve at the time instant 't'. Let A t
and B point on the plot corresponds to the time t and t + O t t + ∆t time
∆t during the motion. As ∆ t approaches zero, the chord AB
becomes the tangent AC at A. The slope of the tangent Instantaneous speed is equal to the slope
of the tangent at given instant.
equal ds/dt, which is equal to the intantaneous speed at
't'.
DC ds
v = tanθ = =
AC dt
The magnitude of average velocity |vavg| and average speed vavg may not be equal, but magnitude of
instantaneous velocity |v| is always equal to instantaneous speed v.
Ex.1 In 1.0 sec a particle goes from point A to point B moving in a semicircle of radius 1.0 m. The
magnitude of average velocity is
(A) 3.14 m/sec (B) 2.0 m/sec (C) 1.0 m/sec (D) zero
Total displacement d A
Sol. Average velocity = = 1m
Total time t o
D = AO + OB 1m
= 1 + 1 = 2m B
t = 1 sec (given)
2
⇒ mg of v of = 2m/sec
1
Ex.3 A body travels the first half of the total distance with velocity v1 and the second half with
velocity v2. Calculate the average velocity :
Sol. Let total distance = 2x. Then
x x v1 + v 2 2x 2v1v 2
total time taken = v + v = x v v ∴ Average speed = = v +v
1 2 1 2 v1 + v 2 1 2
x
v1v 2
(G-1) When velocity is given as a function of t :
Ex.4 Velocity-time equation of a particle moving in a straight line is,
v = (10 + 2t + 3t2)
Find :
(a) displacement of particle from the origin of time t = 1 s, if it is given that displacement is 20 m at
time t = 0
(b) acceleration-time equation.
Sol. (a) The given equation can be written as,
ds
v= = (10 + 2t + 3t 2 )
dt
ds = (10 + 2t + 3t2) dt
s t
or ∫
20
∫
ds = (10 + 2t + 3t 2 )dt
0
or s – 20 = [10t + t2 + t3]01
or s = 20 + 12 = 32 m
(b) Acceleration-time equation can be obtained by differentiating the given equation w.r.t. time.
Thus,
dv d
a= = (10 + 2t + 3 t 2 ) or a = 2 + 6t
dt dt
SPECIMEN PROBLEM :
(A) WHEN EQUATION OF DISPLACEMENT IS GIVEN AND SPEED TO BE FIND OUT
(B) WHEN VELOCITY IS GIVEN AS A FUNCTION OF TIME AND DISTANCE TO BE FIND OUT
In this type of question first find out at what instant the velocity is zero. If this instant is come in our
time limit then distance can be calculated by breaking the integration in two part with modulas
Ex.6 If velocity is depend on time such that v = 4 – 2t. Find out distance travelled by particle from 1
to 3 sec.
Sol. Velocity is zero (4 – 2t = 0) at t = 2 sec
2 3
dx
So for distance
dt
= 4 – 2t ⇒ dx = ∫ (4 – 2t)dt + ∫ (4 – 2t)dt
2
1
dx = 1 + 1 = 2m
1 1
⇒ – +1=t⇒x=
x 1– t
Ex.8 if a = 2x ; initially particle is at x = 2m and is moving with 3 ms–1. Then find out v at x = 5 m.
Sol. Given a = 2x
v 5
vdv
⇒
dx
= 2x ⇒ ∫
3
vdv = ∫ 2xdx
2
v2 9
⇒ – = 25 – 4 ⇒ v2 – 9 = 21 × 2 ⇒ v = 51 ms–1
2 2
(b) Given a = v
v 3
⇒
vdv
dx
=v ⇒ ∫
1
∫
dv = dx
1
⇒ v–1=2 ⇒ v = 3 ms –1
Ex.10 The acceleration of a particle which is depend on time is given by following function
a = 2t + 1
and at time t = 0, x = 1m and u i = 2m/s.
Then find out displacement of the particle at t = 3 sec.
dv dv
Sol. ∴ We know that a = ⇒ = 2t + 1 ⇒ dv = (2t + 1) dt
dt dt
vf t
∫ dv = ∫ (2t + 1)dt
0
2
v f – 2 = t2 + t ⇒ v f = t2 + t + 2
dx
Now v=
dt
xf t
⇒
dx
dt
= t2 + t + 2 ⇒ ∫
1
∫
dx = ( t 2 + t + 2)
0
t3 t2
xf = + + 2t + 1
3 2
So, xf at t = 3 sec is
(3) 3 (3) 2
= + + 2(3) + 1 ⇒ 20.5 m
3 2
So, after t = 3 sec the position of the particle is 20.5m but the displacement of the particle is
= 20.5 – 1 = 19.5 m
Deduce the following equations for unifromly accelerated motion by using intergration technique.
1
(A) v = u + at (B) s = ut + at2
2
a
(C) v2 – u2 = 2as (D) snth = u + (2n – 1)
2
or v – u = a(t – 0)
or v = u + at ...(2)
Second equation of motion. Velocity is defined as
ds
v=
dt
or ds = v dt = (u + at) dt ...(iii)
When time = 0, displacement travelled = 0
When time = t, displacement travelled = s (say).
Integrating equation (3) within the above limits of time and distance, we get
s t t t t
t2
∫ ds =∫ (u + at) dt = u ∫ dt + a ∫ t dt
0 0 0 0
or [s]0s = u[t]0t + a
2 0
t2
or s – 0 = u (t – 0) + a 2 − 0
1 2
or at
s = ut + ...(4)
2
Third equation of motion. By the definitions of acceleration and velocity,
dv dv ds dv
a=
= × = ×v
dt ds dt ds
or ads = vdv ...(5)
When time = 0, velocity = u, displacement travelled = 0
When time = t, velocity = v, displacement travelled = s
(say)
Integrating equation (5) within the above limits of velocity and displacement, we get
s v s v v
v2
∫ a ds = ∫ v dv
0 u
or a ds = ∫
0
∫ v dv
u
or a[s]0s =
2 u
v2 u2
or a[s − 0] =
− or 2as = v2 – u2
2 2
or v2 – u2 = 2as ...(6)
Fourth equation of motion. By definition of velocity,
ds
v=
dt
or ds = vdt = (u + at) dt ...(7)
When time = (n – 1) second, displacement travelled
= sn – 1 (say).
When time = n second, displacement travelled = sn
(say)
Integrating equation (7) within the above limits of time and distance, we get
sn n n n n
t2
∫
sn −1
ds = ∫
n −1
(u + at)dt or [s]ssn = u
n −1 ∫ dt + a ∫t dt or sn − sn −1 = u[t]nn −1 + a
2 n −1
n −1 n −1
a a
= u[n –(n – 1)] + [n2 – (n – 1)2] = u + [n2 – (n2 – 2n + 1)]
2 2
a
snth = u +
(2n − 1) ...(8)
2
where snth = sn – sn –1 = displacement in nth second.
Ex.11 A car starts from rest and accelerates uniformly for 20 seconds to a velocity of 72 km h–1. It
then runs at constant velocity and finally brought to rest in 200 m with a constant retardation.
The total distance covered is 600 m. Find the acceleration, retardation and the total time taken.
Sol. (i) Motion with uniform acceleration
5
Here, u = 0 ; t1 = 20 sec ; v = 72 × = 20 ms–1
18
∴ v = u + at1
20 = 0 + a × 20 or a = 1 m s–2
Distance travelled by car in this time (20 sec),
1 2 1
S1 = ut + at = 0 + × 1 × (20)2 = 200 m
2 2
(ii) Motion with uniform velocity.
As given, total distance = 600 m
we have calculated S1 = 200 m (with uniform acc.)
and S2 = 200 m (with retardation)
∴ Net distance for which body moves with uniform velocity,
S = 600 – S1 – S2
= 600 – 200 – 200 = 200 m
dis tan ce 200
∴ Time taken, t = = = 10 sec.
uniform velocity 20
∴ Total time of journey, t = (20 + 10 + 20) sec
t = 50 sec
Total displacement 600
Average velocity = = = 12 m/s .
Total Time 50
(iii) Motion with uniform retardation.
For this motion, initial velocity, u = 20 m s–1 and final velocity v = 0 ; S2 = 200 m
Acceleration a' = ?
Using, v2 – u2 = 2 a' S2
(0)2 – (20)2 = 2(a′) × 200
a′ = – 1 ms–2
Let t′ = time for which the body comes to rest.
∴ v = u + a′ t ′
0 = 20 – 1t′
∴ t′ = 20 sec.
C. SPECIMEN PROBLEM 2
a=2m/s
Ex.12 Find out distance travelled by the block u=10 m/s
in 10 sec. for a given situation.
Sol. First find out it what instant velocity of block becomes zero.
v = u + at
given : u = 10 m/s, a = – 2m/s2
⇒ 0 = 10 – 2t ⇒ t = 5 sec
u2
cover in deaccelerated motion during t1 to t2 = Total distance = u(∆t) +
2a
Ex.13 Assume that a car is able to stop with a retardation of 8 ms–2 and that a driver can react to an
emergency in 0.5 sec. Calculate the overall stopping distance of the car for a speed of 60 km–1
of the car.
5 50
Sol. Here, u = 60 km h–1 = 60 × = ms−1
18 3
50
Since the application of brakes takes 0.5 s, before this the car was moving with uniform speed of ms −1 .
3
∴ Distance covered in 0.5 sec, with a uniform speed is
50 25
S1 = u × t = × 0.5 = m = 8.33 m
3 3
Now car begins to move with a retardation of 8ms–2
∴ Distance covered before coming to rest,
2a S2 = v2 – u2
2
50
2 2 0− 2500
or v −u 3 = 50 × 50 = = 17.36 m
S2 = = 144
2a −2 × 8 9×2×8
∴ Total (overall) distance = S1 + S2 = 8.33 + 17.36 ⇒ S = 25.69 m
Ex.14 Two buses A and B are at positions 50 m and 100 m from the origin at time t = 0. They start
moving in the same direction simultaneously with uniform velocity of 10 ms–1 and 5 ms–1.
Determine the time and postion at which A overtakes B.
Sol. Here we use equation of motion for constant velocity in Cartesian form.
Given x1 (0) = 50 m, x2 (0) = 100 m,
v1 = 10 ms–1, v2 = 5 ms–1
The positions of the two buses at any instant t are
x1 (t) = x1 (0) + v1t = 50 + 10 t
x2 (t) = x2 (0) + v2t = 100 + 10 t
When A overtakes B,
x1 (t) = x2 (t)
50 + 10t = 100 + 5t or 5t = 50
t = 10 s
x1 (10) = x2 (10) = 150 m
Thus A overtakes B at a position of 150 m from the origin at time t = 10 s.
Ex.15 A bus starts from rest with constant acceleration of 5 ms–2. At the same time a car travelling
with a constant velocity of 50 ms–1 overtakes and passes the bus. (i) Find at what distance will
the bus overtake the car ? (ii) How fast will the bus be travelling then ?
Sol. (i) Suppose the bus overtakes the car after covering distance s.
When the two meet, time taken t is same.
1 2 1 2
For bus, s = ut + at = 0 + × 5 t
2 2
For car, s = 50 t
5 2
∴ t = 50 t or t = 20 s
2
Hence s = 50 t = 50 × 20 = 1000 m.
(ii) v2 = u2 + 2as = 0 + 2 × 5 × 1000 = 10,000 or v = 100 ms–1
SPECIMEN PROBLEM
(E) Maximum Separation :
u=0
a=4m/s2
Ex.16 40 m/s
u2
2u ± 4u2 – 4gH 2u ± 4u 2 – 4g × u2
⇒ t= ⇒t= 2g ∵ Hmax =
2g 2g
2g
u(2 ± 2 )
t= ...(1)
2g
Equation 1 gives two value of time which corresponds to
u(2 – 2 )
t1 = (from ground to Hmax/2 in upward motion)
2g
u(2 + 2 )
t2 = (from ground to Hmax/2 in downward motion)
2g
2u ± 4u 2 – 8gh
⇒ t= h
2g
u
u – u2 – 2gh u + u 2 – 2gh
So, t1 = , t2 = A
g g
⇒ t1 + t2 = T (Time of flight)
1 2 2H
⇒ – H = (0)T – gt ⇒ T =
2 g
(b) Final Velocity when body reaches the ground
from v2 – u2 = 2as
s=–H v = vf u = 0 a = – g
⇒ vf
2
– 0 = 2 (–g) (–H) ⇒ vf = 2gH
Ex.17 A ball is thrown vertically upwards with a velocity u from the ground. The ball allains a maximum
height Hmax. Then find out the time and displacement at which ball have half of the maximum
speed.
Sol. Maximum speed of the ball is u
At point B and C ball have speed u/2 but direction
is opposite so from
B C
v = u + at
u/2 u/2
Let t1 is the time taken by the ball from point A to B and t2 is
the time taken by the ball from A to C h
u
From A to B = u – gt1 ...(i)
2 A
u
From A to C – = u – gt 2 ....(2)
2
u 3u
from (i) t1 = , from (ii) t2 =
2g 2g
from equation v2 – u2 = 2as
2
u 2
⇒ v = ± u/2, u = u, a = – g ⇒ – u = – 2gh
2
u
2
3u 2 3
h= = Hmax ∴ h= Hmax
8g 2g 4
Ex.18 A ball thrown vertically upwards with a speed of 19.6 ms–1 from the top of a tower returns to
the earth in 6 s. Find the height of the tower.
Sol. Here u = 19.6 ms–1
g = –9.8 ms–2
Net displacement, s = – h
Negative sign is taken because displacement is in
the opposite direction of initial velocity.
Tower
1 2
As s = ut + gt h
2
1
∴ – h = 19.6 × 6 + × (–9.8) × 62
2
= 117.6 – 176.4 = –58.8
or h = 58.8 m
Ex.19 A ball is thrown vertically upwards with a velocity of 20 ms–1 from the top of a multistoreyed
building. The height of the point from where the ball is thrown is 25 m from the ground. (i) How
high will the ball rise and (ii) how long will it be before the ball hits the ground?
Sol. (i) Here u = +20 ms–1, g = –10 ms–2
At the highest point, v = 0
Suppose the ball rises to the height h from the point of projection.
As v2 – u2 = 2gs
∴ 02 – 202 = 2 × (–10) × h or h = + 20 m.
(ii) Net displacement, s = –25 m
Negative sign is taken because displacement is in the opposite direction of initial velocity.
1 2
As s = ut + gt
2
1
∴ –25 = 20t + × (–10) × t2
2
or 5t2 – 20t – 25 = 0 or t2 – 4t – 5 = 0
or (t+ 1) (t – 5) = 0
As t ≠ –1, so t = 5s.
Ex.20 A ball thrown up is caught by the thrower after 4s. How high did it go and with what velocity was it
thrown ? How far was it below the highest point 3 s after it was thrown?
Sol. As time of ascent = time of descent
∴ Time taken by the ball to reach the highest point = 2 s
For upward motion of the ball : u = ?, v = 0, t = 2s, g = – 9.8 ms–2
As v = u + gt
∴ 0 = u – 9.8 × 2
or u = 19.6 ms–1
Maximum height attained by the ball is given by
1 2 1
s = ut + gt = 19.6 × 2 + × (9.8) × 22 = 19.6 m.
2 2
Displacement of the ball in 3 s,
1
s = 19.6 × 3 + × (–9.8) × 32 = 58.8 – 44.1 = 14.7 m
2
Distance of the ball from the highest point 3 s after it was thrown
= 19.6 – 14.7 = 4.9 m.
Ex.21 A balloon is ascending at the rate of 9.8 ms–1 at a height of 39.2 m above the ground when a
food packet is dropped from the balloon. After how much time and with what velocity does it reach
the ground?
Take g = 9.8 ms–2.
Sol. Initially the food packet attains the upward velocity of the balloon, so
u = 9.8 ms–1, g = 9.8 ms–2 , s = –39.2 m
Here s is taken negative because it is in the opposite direction of initial velocity.
1 2
Using, s = ut + gt , we get
2
1
– 39.2 = 9.8 t – × 9.8 t2
2
or 4.9 t2 – 9.8 t – 39.2 = 0 or t2 – 2t – 8 =0
or (t – 4) (t + 2) = 0 or t = 4 s or – 2 s
As time is never negative, so t = 4s.
Velocity with which the food packet reaches the ground is
v = u + gt = 9.8 – 9.8 × 4 = – 29.4 ms–1.
Negative sign shows that the velocity is directed vertically downwards.
When a particle is dropped then it will automatically attains the velocity of the frame at that time.
Ex.22 Two balls are thrown simultaneously, A vertically upwards with a speed of 20 ms–1 from the
ground, and B vertically downwards from a height of 40 m with the same speed and along the
same line of motion. At what points do the two balls collide? Take g = 9.8 ms–2.
Sol. Suppose the two balls meet at a height of x from the ground after time t s from the start.
For upward motion of balls A :
u = 20 ms–1, g = – 9.8 ms–2 u=20 ms–1 B
1 2
s = ut + gt
2 40–x
40 m
1
x = 20 t – × 9.8 t2 = 20t – 4.9 t2 ...(i) C
2
For downward motion of ball B, x
1
40 – x = 20 × t + × 9.8 t2
2 u=20 ms–1 A
= 20t + 4.9 t 2
... (ii)
Adding (i) and (ii), 40 = 40 t or t = 1 s
From (i), x = 20 × 1 – 4.9 × (1)2 = 15.1 m
Hence the two balls will collide after 1 s at a height of 15.1 m from the ground.
3. GRAPHS :
2
y=4x +3x
e.g. y = 4 x2 + 3x 2
y=–4x +3x
displacement x2 – x1 B
= = t –t x2
time taken 2 1
x2–x1
= tan θ = slope of the chord AB
x1 A θ
x 2 – x1 t2–t1
vinstantaneous = as lim
t2 → t1 t 2 – t1 t1 t2
when t2 approaches t1 point B approaches Point A and the chord AB becomes tangent to the curve.
Therefore
vinstantaneous = Slope of the tangent x – t curve
x0
(1) Body is at rest at x0.
t
x
(2) Body starts from origin and is moving with speed tan θ away from origin.
θ t
x
(3) Body starts from rest from origin and moves away from origin with increasing
speed velocity and positive acceleration.
x
(4) Body starts from rest from x = x0 and moves away from origin with increasing
velocity or +ve acceleration.
x0
(5) x0 Body starts from x = x0 and is moving toward the origin with constant velocity
passes throw origin after same time and continues to move away from origin.
x
x0
(6) Body starts from rest at x = x0 and then moves with increasing speed towards
origin
∴ acceleration is –ve
t
x
(7) Body starts moving away from origin with some initial speed. Speed of body is
decreasing till t1 and it becomes 0 momentarily of t = t1 and At this instant. Its
reverses its direction and move towards the origin with increasing speed.
t2
t
O t1
(8) x Body starts from origin moves away from origin in the –ve x-axis at t = t1 with
decreasing speed and at t= t1 it comes at rest momentarily, Reverses its direction
t1 moves towards the origin the increasing speed. Crosses the origin at t = t2.
t
t2
x
(9) Body starts from origin from rest and moves away from origin with increasing
t speed.
t
v
t
v
(3) Body is at rest initially then it starts moving with its velocity increasing at a
constant rate i.e. body is moving with constant acceleration.
v t
(4) Body starts its motion with initial velocity v0 and continues to move with its
v0 velocity increasing at a constant rate i.e. acceleration of the body is constant.
t
v
(5) Body starts its motion with initial velocity v0. Then it continues to move with its
velocity decreasing at a constant rate i.e. acceleration of the body is negative
v0
and constant. At t = t0 the body comes to rest instantaneously and reverses its
t0 direction of motion and then continues to move with decreasing velocity or increasing
t speed.
For 0 < t < t0 motion of the body is deaccelerated (∴ speed is decreassing)
t > t0 motion of the body is accelerated (∴ speed is increasing)
v
(6) Body is at rest initially. Then it starts moving with increasing velocity. As time
increases its velocity is increasing more rapidly. i.e. the moving with increasing
acceleration.
t
v
(7) v0 Body starts its motion with initial velocity v0. Its velocity is decreasing with
time and at t = t0 . It becomes zero after body reverse its direction of motion and
continues to move with decreasing velocity or increasing speed. Since velocity
of the body is decreasing for whole motion. Therefore, its acceleration is
t0 t negative.For 0 < t < t0 motion of the body is deaccelerated (speed is
decreassing) t > t0 motion of the body is accelerated (∵ speed is increasing)
(1) acceleration of the body is zero that means the body is moving constant velocity.
t
(3) Acceleration of the body is constant and negative
(4) Initially the acceleration of the body is zero. Then its acceleration is increasing
at a constant rate.
t
(6) Initially acceleration of the body is zero. Its acceleration is positive for whole of
its motion. Its acceleration is increasing for whole of its motion.
t
t
v0
0
u
x=
t t
(ii) If at t = 0, x = x0 then
x v
+ v 0t
x0 x 0
v0
x=
t t
(iii) If at t = 0, x = – x0 then
x v
t t
+v
0
v0
–x
0
x=
t t
–x0
a0
slope = tanθ
= a0
θ
t t
v = a0 t
(ii) If u = u0 , a = a0
1
x = xi + u0t + a0 t 2 v = u0 + a0t
2
x x
v a
a0
if xi = 0 t if xi = x0 t t t
(iii) if u = u0, a = – a0
1 2
x = xi + u0t – a 0 t
2
x
x
x0
if xi = 0 if xi = x0
t t
t0
v
a
u0
t
t0
t –a0
(iv) if u = – u0 , a = + a0
1
x = xi – u0t + a0 t 2
2
x
x0
if xi = 0 if xi = x0
v
a
a0
t
–u0 t
(v) If u = u0, a = – a0
1
x = xi – u0t – a0 t 2
2
x x
x0
t t
if xi = 0 if xi = x0
v
a
t t
–v0
–a0
(ii) If a body is dropped from a height h above the ground. Take dropping point to be origin and
upward direction as +ve.
1 2
x= – gt
2
x v a=–g
a
2h 2h
g g
t t
t
v = – gt
–h – 2gh –g
(iii) If a body is projected vertically upwards from a tower of height h with initial velocity u. Take
the projection point to be origin and upward direction as +ve.
x
u2 v
2g a
2u
g
t u/2g t
u v = u – gt t
g
–g
–h
(iv) A car starting from rest accelerates uniformly at 2 ms–2 for 5 seconds and then moves with
constant speed acquired for the next 5 seconds and then comes to rest retarding at 2 ms–2.
Draw its
(a) Position vs time graph
(b) Velocity vs time graph
(c) acceleration vs time graph
(v) A particle starts from x = 0 and initial speed 10 ms–1 and moves with constant speed 10ms–1 for
20 sec. and then retarding uniformly comes to rest in next 10 seconds.
acceleration vs time graph
a
v
–1
10ms
20 30
t (sec)
–2 t (sec)
–1ms 20 30
Acceleration vs time graph velocity vs time graph
x
250m
200m
t (sec)
20 30sec
(V) Conversion of velocity v/s time graph to speed v/s time graph.
As we know that magnitude of velocity represent speed therefore whenever velocity goes –ve take its
mirror image about time axis.
velocity
speed
e
ag
i m
or
Ex-26 irr
m
t (sec)
t (sec)
velocity speed e
ag
im
or
irr
Ex-27 m
t t
Dist./Displacement
For distance time graph just make the mirror image of the displacement
Dist.-time
time graph from point of zero velocity onwards. C
v0
t0
v
(i) ⇒
x=
t t
v x a
a0
0
a
=
nθ
(ii) ⇒ a - t graph
ta
t t t
v x
a
tan θ = – a0
t
(iii) ⇒
θ
t0 t
t –a0
t0
at t = t0 velocity reverses its direction.
v
(iv) x – t graph
From t = 0 to t = t1 acceleration = 0 therefore v0
from t = 0 to t = t1, x - t graph will be a straight line.
From t = t1 to t2 acceleration is negative
∴ It will be an opening downward parabola
x t1 t2 t
t1 t2 t
v
(v) upto t = t1 acceleration is +ve
t1 < t < t2 acceleration is zero.
t > t2 acceleration is –ve
x t
t1 t2
x - t graph
t1 t2 t
1
vt = 30 sec = 154 + × 10 × 10 = 204 ms–1
2
(VIII) Reading of graphs if the motion of two bodies are sketched on the same axes.
(a) Reading of x - t graphs
x
x3
B
x2
x1
A
O t1 t2 t3 t
Conclusions :
(i) Body A Start its motion at t = 0 from origin and is moving away from the origin with constant velocity.
Finally it ends its motion at a distance of x2m from origin at t = t3.
(ii) Body B starts its motion at t = t1 from origin and is moving away from origin with constant velocity.
Finally it ends its motion at a distance of x3m from origin at t = t3
(iii) Since slope of B is greater than slope of A. Therefore velocity of B is greater than velocity of A.
(iv) A t = t2, Both A & B are at the same distance from starting point that means B overtakes A at t = t2
(v) ∵ velocity of both A & B are constant
∵ acceleration of both the bodies are zero.
(vi) ∴ x3 > x2
∴ At the end of the motion B is at a greater distance from the starting point.
x2 A
B
x1
Ex-31 x0
t0 t1 t
Conclusion :
(i) Body A starts its motion at t = 0 from origin and is moving away from the origin with constant
velocity. Finally its motion ends at t = t1 at x = x2 m.
(ii) Body B starts its motion at t = 0 from x = x0 and then moves with constant velocity away from the
origin. Finally it ends its motion at t = t1.
(iii) Velocity of A is greater than that of B.
(iv) At t = t0 A overtakes B
(v) acceleration of both A & B is zero.
(vi) ∵ x2 > x1
∴ At the end of the motion A is at a greater distance from the starting point then B
x
B
A
Ex-32
t1 t2 t
Conclusions :
(i) Both A & B starts their motion at same time t = 0 and from same point x = 0.
(ii) Both are moving away from the starting point.
(iii) A is moving with constant velocity while B starts its motion from rest and its velocity is increasing
with time i.e. it has some positive acceleration.
(iv) ∵ At t = t1 the tangent on B's graph becomes parallel to the A's graphs
∴ At t = t1 velocity of both A & B is same.
(v) For t < t1 velocity of A is greater than velocity of B. Therefore up to t = t1, separation between A
& B increases with time.
(vi) For t > t1 velocity of B is greater than velocity of A. Therefore after t = t1 separation between A &
B starts decreasing and it becomes zero at t = t2 where B overtakes A.
Y u
⇒ usinθ +
u cos θ
θ
O x
Assume that effect of air friction and wind resistance are negligible and value of ‘acceleration due to
→
gravity g is constant.
Take point of projection as origin and horizontal and vertical direction as +ve X and Y-axes, respectively.
For X-axis For Y - axis
ux = u cosθ, uy = u sinθ
ax = 0, ay = – g,
vx = u cosθ, and vy = u sinθ – gt, and
1
x = u cosθ × t y = u sinθ t – gt2
2
It is clear from above equations that horizontal component of velocity of the particle remains constant
while vertical component of velocity is first decreasing, gets zero at the highest point of trajectory and
then increases in the opposite direction. At the highest point, speed of the particle is minimum.
The time, which projectile takes to come back to same (initial) level is called the time of flight (T).
At initial and final points, y = 0,
1
So u sinθ t – gt2 = 0
2
2u sin θ 2u sin θ
⇒ t = 0 and t = So, T=
g g
Range (R) The horizontal distance covered by the projectile during its motion is said to be range of the
projectile
u 2 sin 2θ
R = u cosθ × T =
g
For a given projection speed, the range would be maximum for θ = 45°.
Maximum height attained by the projectile is
u 2 sin 2 θ
H=
2g
at maximum height the vertical component of velocity is 0.
u sin θ T
Time of ascent = Time of descent = =
g 2
Speed, kinetic energy, momentum of the particle initialy decreases in a projectile motion and attains a
θ is the angle between v and horizontal which decreases to zero. (at top most point) and again
Ex.33 A body is projected with a velocity of 30 ms–1 at an angle of 30° with the vertical. Find the
maximum height, time of flight and the horizontal range.
Sol. Here u = 30 ms–1,
Angle of projection, θ = 90 – 30 = 60°
Maximum height,
u 2 sin 2 θ 30 2 sin 2 60°
H= = = 34.44 m
2g 2 × 9.8
Time fo flight,
2u sin θ 2 × 30 sin 60°
T= = = 5.3 s
g 9.8
Horizontal range,
u 2 sin 2θ 30° sin 120° 30 2 sin 60°
R= = = = 79.53 m.
g 9.8 9.8
Ex.34 Find out the relation between uA, uB, uC (where uA, uB, uC are the initial velocities of particles A,
B, C, respectively)
B C
A
2u y
⇒ TA = TB = TC g
2u xu y
from figure RC > RB > RA ∵R =
g
⇒ uxC > uxB > uxA ⇒ uA < uB< uC
vy u sin θ – gt –1 u sin θ – gt
tan α = = ⇒ α = tan
vx u cos θ u cos θ
v= u2 – 2gh
Note that this is the velocity that a particle would have at height h if it is projected vertically from
ground with u.
Ex.35 A body is projected with a velocity of 20 ms–1 in a direction making an angle of 60° with the
horizontal. Calculate its (i) position after 0.5 s and (ii) velocity after 0.5 s.
Sol. Here u = 20 ms–1, θ = 60° , t = 0.5 s
(i) x = (u cosθ)t = (20 cos60°) × 0.5 = 5 m
1 2
y = (u sin θ) t – gt = (20 × sin 60°) × 0.5
2
1
– × 9.8 × (0.5)2 = 7.43 m
2
(ii) vx = u cos θ = 20 cos 60° = 10 ms–1
vy = u sin θ – gt = 20 sin 60° – 9.8 × 0.5
= 12.42 ms–1
vy 12.42
∴ v= v 2x + v 2y = (10) 2 + (12.42) 2 =15.95 ms
–1
tan β =
vx
= = 1.242
10
∴ β = tan–1 1.242 = 51.16°.
usinθ P(x,y)
Max.
u y height=h
θ
m
vx=u cosθ
O ucosθ X
B θ
R v
uy
∵ The horizontal distance covered by the body in time t,
x = Horizontal velocity × time = u cos θ. t
x
or t =
ucos θ
For vertical motion : u = u sinθ, a = –g, so the vertical distance covered in time t is given by
1 2 x 1 x2
s = ut + at or y = u sin θ. – g. 2
2 u cos θ 2 u cos 2 θ
1 x2
or y = x tanθ – g 2 ...(1)
2 u cos 2 θ
gx cos θ gx
y = x tan θ 1 – 2 2 ⇒ y = x tan θ 1 – 2u 2 cos θ sin θ
2u cos θ sin θ
x
y = x tan θ 1 – ...(2)
R
Ex.36 A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and falls
at a distance of 14 m from the wall. Find the magnitude and direction of the velocity.
Sol. The ball passes through the point P(4, 4). So its range = 4 + 14 = 18m.
The trajectory of the ball is,
Now x = 4m, y = 4m and R = 18 m
y
4 7
∴ 4 = 4 tan θ 1 – = 4 tanθ . P(4,4)
18 9 u
9 9 7
or tan θ = , sin θ = , cosθ = 4m
7 130 130
θ
18 × 9.8 × 130
or u = 2
= 182 4m x
2×9×7 14m
or u = 182 = 13.5 ms
–1
Ex.37 A particle is projected over a triangle from one end of a horizontal base and grazing the vertex
falls on the other end of the base. If α and β be the base angles and θ the angle of projection,
prove that tan θ = tan α + tan β.
Sol. If R is the range of the particle, then from the figure we have
y y y(R – x) + xy
tan α + tan β = + = Y
x R– x x(R – x)
y R
or tanα + tan β = × ...(1) P(x,y)
x (R – x)
Also, the trajectrory of the particle is
y
θ
x α β
y = x tan θ 1– O
x
R x B A
R–x
y R
or tanθ = ×
x (R – x)
From equations (1) and (2), we get
tan θ = tan α + tan β .
1 2h
or h = 0 × T + gT2 or T=
2 g
Horizontal range. It is the horizontal distance covered by the projectile during its time of flight. It is
equal to OA = R. Thus R = Horizontal velocity × time of flight = u × T
2h
or R=u
g
Velocity of the projectile at any instant. At the instant t (when the body is at point P), let the
velocity of the projectile be v. The velocity v has two rectangular components:
Horizontal component of velocity, vx = u
Vertical component of velocity, vy = 0 + gt = gt
∴ The resultant velocity at point P is
v = v2x + v2y = u2 + g2 t2
If the velocity v makes an angle β with the horizontal, then
vy gt gt
tan β = = or β = tan–1
vx u u
Ex.38 A body is thrown horizontally from the top of a tower and strikes the ground after three seconds
at an angle of 45° with the horizontal. Find the height of the tower and the speed with which
the body was projected. Take g = 9.8 ms–2.
Sol. As shown in figure, suppose the body is thrown horizontally from the top O of a tower of height y with
velocity u. The body hits the ground after 3s. Considering verticlly downward motion of the body,
1 2 1
y = uyt + gt = 0 × 3 + ×9.8 × (3)2 = 44.1 m [∴ Initial vertical velocity, uy = 0]
2 2
Final vertical velocity,
vy = uy + gt = 0 + 9.8 × 3 = 29.4 ms–1
Final horizontal velocity, vx = u
As the resultant velocity u makes an angle of 45° with the horizontal, so
vy 29.4
tan 45° = or 1 = or u = 29.4 ms–1.
vx x
Ex.39 A particle is projected horizontally with a speed u from the top of plane inclined at an angle θ
with the horizontal. How far from the point of projection will the particle strike the plane?
Sol. The horizontal distance covered in time t,
x
x = ut or t =
u u
The vertical distance covered in time t, θ
1 2 1 x2
y=0+ gt = g × 2 [using (1)] y D
2 2 u
y gx 2 θ
Also = tan θ or y = x tan θ ∴ = x tan θ
x 2u 2 x=ut
gx
or x 2 – tan θ = 0
2u
2u 2 tan θ
As x = 0 is not possible, so x =
g
The distance of the point of strike from the point of projection is
D= x2 + y2 = x2 + (x tan θ)2
2u2
=x 1 + tan2 θ = x sec θ or D = tan θ sec θ
g
Ex.40 A ball rolls off the top of a stairway with a constant horizontal velocity u. If the steps are h
2hu 2
metre high and w meter wide, show that the ball will just hit the edge of nth step if n =
gw 2
Sol. Refer to figure. For n th step,
net vertical displacement = nh u
net horizontal displacement = nω 1st
Let t be the time taken by the ball to reach the nth step. Then 2nd
R = ut h
nω w
or nω = ut or t=
u
1 2
Also, y = uy t + gt nth
2 R
2
1 2 1 nω 2hu2
or nh = 0 + gt = g or n =
2 2 u gω2
y x
B
u
gsinβ
β gcosβ
α
β g
β O
O
C
Now, let us derive the expressions for time of flight (T) and range (R) along the plane.
Time of flight
1
At point B displacement along y-direction is zero. So, substituting the proper values in sy = uyt + ay t2 ,
2
we get
1 2u sin α
α+ (– g cos β ) t2 ∴ t = 0 and
g cos β
0 = u t s i n
2
2u sin α
t = 0, corresponds to point O and t = corresponds to point B. Thus,
g cos β
2u sin α
T=
g cos β
Range
Range (R) or the distance OB is also equal to be displacement of projectile along x-direction in the
t = T. Therefore.
1 1
R = sa = uxT + axT2 ⇒ R = u cos α T – sin β T2
2 2
Ex.42 A particle is projected at an angle α with horizontal from the foot of a plane whose inclination to
horizontal is β . Show that it will strike the plane at right angles if cotβ = 2 tan (α – β)
Sol. Let u be the velocity of projection so that u cos (α – β ) and u sin (α – β ) are the initial velocities
respectively parallel and perpendicular to the inclined plane. The acceleration in these two directions
are (–g sin β ) and (–g cos β ).
The initial component of velocity perpendicular to PQ is u sin (α – β ) and the acceleration in this
direction is (–g cosβ ). If T is the time the particle takes to go from P to Q then in time T the space
described in a direction perpendicular to PQ is zero.
1 u
0 = u sin (α – β ).T – g cos β .T2 Q
2
2u sin(α – β)
T=
g cos β
α
If the direction of motion at the instant when the particle β
hits the plane be perpendicular to the plane, then the
velocity at that instant parallel to the plane must be zero. P N
∴ u cos (α – β ) – g sin β T = 0
u cos(α – β) 2u sin(α – β)
=T=
g sin β g cos β
∴ cosβ = 2 tan (α – β )
Ex.43 Two inclined planes OA and OB having inclinations 30° and 60° with x
y
the horizontal respectively intersect each other at O, as shwon in
u v B
figure. a particle is projected from point P with velocity u = 10 3 m / s
along a direction perpendicular to plane OA. If the particle strikes A Q
1
Therefore, h = PO sin 30° = (10) or h = 5m Ans.
2
or OQ = 10 3 m
5. RELATIVE MOTION
The word 'relative' is a very general term, which can be applied to physical, nonphysical, scalar or
vector quantities. For example, my height is five feet and six inches while my wife's height is five feet
and four inches. If I ask you how high I am relative to my wife, your answer will be two inches. What
you did? You simply subtracted my wife's height from my height. The same concept is applied everywhere,
whether it is a relative velocity, relative acceleration or anything else. So, from the above discussion
→
we may now conclude that relative velocity of A with respect of B (written as v AB ) is
→ → →
v AB = v A – v B
Similarly, relative acceleration of A with respect of B is
→ → →
a AB = a A – a B
If it is a one dimensional motion we can treat the vectors as scalars just by assigning the positive sign
to one direction and negative to the other. So, in case of a one dimensional motion the above
equations can be written as
vAB = vA – vB
and aAB = aA – aB
Further, we can see that
→ → → →
v AB = – v BA or a BA = – a AB
Ex.44 Seeta is moving due east with a velocity of 1 m/s and Geeta is moving the due west with a
velocity of 2 m/s. What is the velocity of Seeta with respect to Geeta?
Sol. It is a one dimensional motion. So, let us choose the east direction as positive and the west as
negative. Now, given that
vs = velocity of Seeta = 1 m/s
and vG = velocity of Geeta = – 2m/s
Thus, vSG = velocity of Seeta with respect to Geeta
= vS – vG = 1 – (–2) = 3 m/s
Hence, velocity of Seeta with respect to Geeta is 3 m/s due east.
IMPORTANT NOTE :
PROCEDURE TO SOLVE THE VECTOR EQUATION.
A =B+C ...(1)
(a) Their are 6 variables in this equation which are following :
(1) Magnitude of A and its direction
(2) Magnitude of B and its direction
(3) Magnitude of C and its direction.
(b) We can solve this equation if we know the value of 4 varibales [Note : two of them must be directions]
(c) If we know the two direction of any two vectors then we will put them on the same side and other on
the different side.
For example
If we know the directions of A and B and C' s direction is unknown then we make equation as follows : -
C = A –B
(d) Then we make vector diagram according to the equation and resolve the vectors to know the
unknown values.
Ex.45 Car A has an acceleration of 2 m/s2 due east and car B, 4 m/s2 due north. What is the acceleration
of car B with respect to car A?
Sol. It is a two dimensional motion. Therefore, N
→
a BA = acceleration of car B with respect to car A
→ → W E
= aB = – a A
→
Here, a B = acceleration of car
S
B = 4 m/s2 (due north)
→
and a A = acceleration of car A = 2 m/s2 (due east)
→ →
→ a BA a B = 4m / s 2
2 2 2
| a BA |= (4) + (2) = 2 5m / s
4
and α = tan–1 = tan–1(2)
2 α
→ →
Thus, a BA is 2 5 m/s2 at an angle of α = tan–1(2) – a A = 2m / s 2
from west towards north.
Ex.46 Three particle A, B and C situated at the vertices of an equilateral triangle starts moving simul-
taneously at a constant speed "v" in the direction of adjacent particle, which falls ahead in the
anti-clockwise direction. If "a" be the side of the triangle, then find the time when they meet.
A
Sol. Here, particle "A" follows "B", "B" follows "C" and "C" follows
"A". The direction of motion of each particle keeps chang-
ing as motion of each particle is always directed towards
other particle. The situation after a time "t" is shown in the
figure with a possible outline of path followed by the par- O
ticles before they
meet. B C
When two bodies are in motion, the questions like, the minimum distance between them or the time
when one body overtakes the other can be solved easily by the principle of relative motion. In these
type of problems one body is assumed to be at rest and the relative motion of the other body is
considered. By assuming so two body problem is converted into one body problem and the solution
becomes easy. Following example will illustrate the statement.
Ex.47 Car A and car B start moving simultaneously in the same direction along the line joining them.
Car A with a constant acceleration a = 4 m/s2, while car B moves with a constant velocity v = 1
m/s. At time t = 0, car A is 10 m behind car B. Find the time when car A overtakes car B.
Sol. Given : uA = 0, uB = 1 m/s, aA = 4m/s2 and aB = 0
Assuming car B to be at rest, we have
uAB = uA – uB = 0 – 1 = – 1 m/s
aAB = aA – aB = 4 – 0 = 4 m/s2
Now, the problem can be assumed in simplified form as follow :
2 2
a=4m/s v=1m/s
A 10m B
+ve
Substituting the proper values in equation
2
uAB= –1m/s, aAB= 4m/s
A 10m B
At rest
1 2
s = ut + at
2
1
we get 10 = – t + (4)(t2 ) or 2t2 – t – 10 = 0
2
1 ± 1 + 80 1 ± 81 1±9
or t= = = or t = 2.5 s and – 2 s
4 4 4
Ignoring the negative value, the desired
time is 2.5s. Ans.
Note : The above problem can also be solved without using the concept of relative motion as under.
At the time when A overtakes B,
sA = sB + 10
1
∴ × 4 × t 2 = 1 × t + 10
2
or 2t2 – t – 10 = 0
Which on solving gives t = 2.5 s and – 2 s, the same as we found above.
As per my opinion, this approach (by taking absolute values) is more suitable in case of two body
problem in one dimensional motion. Let us see one more example in support of it.
Ex.48 An open lift is moving upwards with velocity 10m/s. It has an upward acceleration of 2m/s2. A
ball is projected upwards with velocity 20 m/s relative to ground. Find :
(a) time when ball again meets the lift.
(b) displacement of lift and ball at that instant.
(c) distance travelled by the ball upto that instant. Take g = 10 m/s2
Sol. (a) At the time when ball again meets the lift,
sL = sB
1 1 2m/s2 10m/s 20m/s
∴ 10t + × 2 × t2 = 20 t – × 10t2
2 2 +ve
Solving this equation, we get Ball
2
5 10m/s
t=0 and t= s
3 L Lift B Ball
5
∴ Ball will again meet the lift after s.
3
(b) At this instant
2
5 1 5 175
sL = sB = 10 × + ×2× = m = 19.4 m
3 2 3
9
(c) For the ball u ↑ ↓a . Therefore, we will first find t0, the time when its velocity becomes zero.
u 20
t0 = = = 2s
a 10
5
As t = s < t0 , distance and displacement are equal
3
or d = 19.4 m Ans.
Concept of relative motion is more useful in two body problem in two (or three) dimensional motion.
This can be understood by the following example.
Ex.49 Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is
streaming west at 20 km/h and ship B is streaming north at 20km/h. What is their distance
of closest approach and how long do they take to reach it ?
Sol. Ships A and B are moving with same speed 20 km/h in
the directions shown in figure. It is a two dimensional, N
two body problem with zero acceleration. Let us find vA A E
vBA
vB
vBA = vB − v A
B
Here, | vBA |= (20)2 + (20)2 AB=10km
= 20 2 km / h
i.e., vBA is 20 2 km / h at an angle of 45º from east
towards north. Thus, the given problem can be simplified as :
45º
A is at rest and B is moving with vBA in the direction shown in figure. A
Therefore, the minimum distance between the two is C
vBA
smin = AC = AB sin 45º 45º
B
1
= 10 km = 5 2 km Ans.
2
and the desired time is
BC 5 2
t= = (BC = AC = 5 2 km )
| vBA | 20 2
1
= h = 15 min Ans.
4
→
A boatman starts from point A on one bank of a river with velocity v br in the direction shown in fig.
→
River is flowing along positive x-direction with velocity v r . Width of the river is w, then
→ → →
v b = v br + v r
Therefore, vbx = vrx + vbrx = vr – vbr sinθ
and vby = vry + vbry
= 0 + vbr cosθ = vbr cosθ
Now, time taken by the boatman to cross the river is :
w w
t= =
v by v br cos θ
w
or t = v cos θ ...(i)
br
Further, displacement along x-axis when he reaches on the other bank (also called drift) is :
w
x = vbx t = (vr – vbr sin θ) v cos θ
br
w
or x = (vr – vbr sinθ) v cos θ ...(ii)
br
Three special are :
(i) Condition when the boatman crosses the river in shortest interval of time
B
From Eq.(i) we can see that time (t) will be minimum when θ = 0°,
i.e., the boatman should steer his boat perpendicular to the river →
current. vbr
w
Also, tmin = v as cos θ = 1 A →
br
vr
(ii) Condition when the boatman wants to reach point B, i.e., at a point just opposite from where
he started
In this case, the drift (x) should be zero.
∴ x=0 B
w →
or (vr – vbr sinθ) v cos θ = 0 v br
br
θ
or vr = vbr sin θ
A →
–1 v
vr vr
or sinθ = v or θ = sin r
br v br
–1 v
Hence, to reach point B the boatman should row at an angle θ = sin r upstream from AB.
v br
Further, since sinθ not greater than 1.
So, if vr ≥ vbr, the boatman can never reach at point B. Because if vr = vbr, sin θ = 1 or θ = 90° and it is
just impossible to reach at B if θ = 90°. Moreover it can be seen that vb = 0 if vr = vbr and θ = 90°.
Similarly, if vr > vbr, sinθ > 1, i.e., no such angle exists. Practically it can be realized in this manner that
it is not possible to reach at B if river velocity (vr) is too high.
(iii) Shortest path
Path length travelled by the boatman when he reaches the opposite shore is
s= w 2 + x2
Here, w = width of river is constant. So for s to be minimum modulus of x (drift) should be minimum.
Now two cases are possible.
–1 v r –1 v r
when θ = sin v or smin = w at θ = sin
br v br
dx
When vr > vbr : In this case x is minimum, where =0
dθ
d w
or (vr – vbr sin θ) = 0
dθ vbr cos θ
or –vbr cos2θ – (vr – vbr sinθ) (– sinθ) = 0
or – vbr + vr sinθ = 0
vbr
or θ = sin–1
vr
Now, at this angle we can find xmin and then smin which comes out to be
vr –1 vbr
smin = w at θ = sin
vbr vr
Ex.50 A man can row a boat with 4 km/h in still water. If he is crossing a river where the current is 2
km/h.
(a) In what direction will his boat be headed, if he wants to reach a point on the other bank, directly
opposite to starting point?
(b) If width of the river is 4 km, how long will the man take to cross the river, with the condition in
part (a)?
(c) In what direction should he head the boat if he wants to cross the river in shortest time and
what is this minimum time?
(d) How long will it take him to row 2 km up the stream and then back to his starting point ?
Sol. (a) Given, that vbr = 4 km/h and vr = 2 km/h
vr 2 1
∴ θ = sin–1 v = sin–1 = sin–1 = 30°
br 4 2
Hence, to reach the point directly opposite to starting point he should head the boat at an angle of
30° with AB or 90° + 30° = 120° with the river flow.
(b) Time taken by the boatman to cross the river
w = width of river = 4 km
vbr = 4 km/h and θ = 30°
4 2
∴ t= = h Ans.
4 cos 30° 3
(c) For shortest time θ = 0°
w 4
and tmin = v cos 0° = = 1h
br 4
Hence, he should head his boat perpendicular to the river current for crossing the river in shortest time
and this shortest time is 1 h.
vbr–vr vbr+vr
D C D C
Ex.51 A man can swim at a speed of 3 km/h in still water. He wants to cross a 500 m wide river
flowing at 2 kh/h. He keeps himself always at an angle of 120° with the river flow while swim-
ming.
(a) Find the time he takes to cross the river.
(b) At what point on the opposite bank will he arrive ?
Sol. The situation is shown in figure
Here vr,g = velocity of the river with respect to the ground
Y
B C
vm,r = velocity of the man with respect to the river
vm,g = velocity of the man with respect to the ground.
,g
vm
(a) We have
°
30
vm,g = vm,r + vr,g ...(i) vm,r = 3km/h
θ
Hence, the velocity with respect to the ground is along AC.
Taking y-components in equation (i),
A vr,g = 2km/h
3 3
vm,g sinθ = 3 km/h cos 30° + 2 km/h cos 90° = km/h
2
Time taken to cross the river
displacement along the Y - axis 1/ 2km 1
= = = h
velocity along the Y - axis 3 3 / 2 km / h 3 3
(b) Taking x-components in equation (i),
1
vm,g cos θ = –3km/h sin 30° + 2 km/h = km / h
2
Displacement along the X-axis as the man crosses the river
= (velocity along the X-axis) (time)
1km 1 1
= × h = km
2h 3 3 6 3
Ex.52 A boat moves relative to water with a velocity v and river is flowing with 2v. At what angle the
boat shall move with the stream to have minimum drift?
(A) 30° (B) 60° (C) 90° (D) 120°
Sol. (D) Let boat move at angle θ to the normal as shown in
1
figure then time to cross the river =
v cos θ
ucos θ
1 ub = u
drift x = (2v – v sin θ) for x to be minimum I = width of river
v cos θ
dx
= 0 = 1 (2 sec θ tan θ – sec2θ) or sin θ = 1/2
dθ
u sinθ ur=2v
or θ = 30° and φ = 90 + 30 = 120°
→ → → → →
and v b is replaced by v a (absolute velocity of aircraft). Further,, v a = v aw + v . The following
w
example will illustrate the theory.
Ex.53 If two vectors A and B make angle 30° and 60° B
with their resultent and B has magnitude equal to
60°
10, then find magnitude of A . 30°
Bsin60° A
So B sin 60° = A sin 30°
⇒ 10 sin 60° = A sin 30° A sin 30°
⇒ A = 10 3
Ex.54 An aircraft flies at 400 km/h in still air. A wind of 200 2 km/h is blowing from the south. The
pilot wishes to travel from A to a point B north east of A. Find the direction he must steer
and time of his journey if AB = 1000 km.
Sol. Given that vw = 200 2 km/h
→ →
vaw = 400 km/h and v a should be along AB or in north-east direction. Thus, the direction of v aw
→ →
should be such as the resultant of v w and v aw is along AB or in north - east direction.
→ N
Let v aw makes an angle α with AB as shown in figure.
B
Applying sine law in triangle ABC, we get
→
AC
=
BC v a 45° →v w = 200 2km / h
sin 45° sin α
BC 200 2 1 1 45°
α → C
or sin α = sin 45° = = A v aw = 400 km / h
AC 400 2 2 E
∴ α = 30°
Therefore, the pilot should steer in a direction at an angle of (45° + α) or 75° from north towards
east.
→
| v a| 400 → sin 105° km
Further, = or | v a | = sin 45° × (400) h
sin(180°–45°–30° ) sin 45°
→
vm = velocity of man (it may be velocity of cyclist or velocity of motorist also)
→
and vrm = velocity of rain with respect to man.
→
Here, v is the velocity of rain which appears to the man. Now, let us take one example of this.
rm
Ex.55 A man standing on a road has to hold his umbrella at 30° with the vertical to keep the rain
away. He throws the umbrella and starts running at 10 km/h. He finds that raindrops are hitting
his head vertically. Find the speed of raindrops with respect to (a) the road, (b) the moving man.
Sol. When the man is at rest with respect to the ground, the rain comes to him at an angle 30° with the
vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation
when the man runs is shown in the figure
30° vm,g
30°
v r,m vr,g
(b)
(a)
Here vr,g = velocity of the rain with respect to the ground
vm,g = velocity of the man with respect to the ground and vr,m = velocity of the rain with respect to
the man.
We have, vr,g = vr,m + vm,g ...(i)
Taking horizontal components, equation (i) gives
10 km / h
vr,g sin30° = um,g = 10 km/h or, v,g = = 20km / h
sin 30°
Taking vertical components, equation (i) gives
3
vr,g cos30° = vr,m or, vr,m = (20 km/h) = 10 √ 3 km/h.
2
Ex.56 To a man walking at the rate of 3 km/h the rain appears to fall vertically. When the increases
his speed to 6 km/h it appears to meet him at an angle of 45° with vertical. Find the speed of rain.
Sol. Let i and j be the unit vectors in horizontal and vertical directions respectively..
Let velocity of rain Vertical ( j )
→
v r = aiˆ + bj
ˆ ...(i)
Then speed of rain will be
→
| v r |= a2 + b2
Horizontal ( i )
→
In the first case v m = velocity of man = 3 i
→ → →
∴ v rm = v r – v m = (a – 3)iˆ + bj
ˆ
It seems to be in vertical direction. Hence,
a – 3 = 0 or a = 3
→
In the second case v m = 6 i
→
∴ ˆ = – 3 i + b j
v rm = (a – 6)iˆ + bj
This seems to be at 45° with vertical.
Hence, |b| = 3
Therefore, from Eq. (ii) speed of rain is
→
| v r |= (3)2 + (3)2 = 3 2 km / h Ans.
Y Y
u1 u2
α1 α2
X X
(A) (B)
i.e., the relative motion between the two particles is uniform. Now
u1x = u1 cos α1, u2x = u2 cos α2
u1y = u1 sin α1 and u2y = u2 sin α2
Therefore, u12x = u1x – u2x = u1 cos α1– u2cos α2 y
and u12y = u1y – u2y = u1 sin α1– u2 sin α2
u12x and u12y are the x and y components of relative
u12y
velocity of 1 with respect to 2.
Hence, relative motion of 1 with respect to 2 is a straight u12
a12=0
θ
u x
line at an angle θ = tan −1 12 y with positive x-axis. u12x
u12 x
Now, if u12x = 0 or u1 cos α1 = u2 cos α2, the relative motion is along y-axis or in vertical direction
(as θ = 90º). Similarly, if u12y = 0 or u1 sin α1 = u2 sin α2, the relative motion is along x-axis or in
horizontal direction (as θ = 0º).
Note : Relative acceleration between two projectiles is zero. Relative motion between them is uniform.
Therefore, condition of collision of two particles in air is that relative velocity of one with respect to
the other should be along line joining them, i.e., if two projecticles A and B collide in mid air, then VAB
should be along AB or VBA along BA.
Condition for collision of two projectiles : Consider the situation shown in the figure. For projectiles to
collide, direction of velocity of A with respect to B has to be along line AB.
Here, vABx = u1 cos α1 + u2 cos α2
u2
vABy = u1 sin α1 – u2 sin α2 B
Y
Let, direction of velocity vector of A(wrt B) is making an u1
h2
angle β with +ve X-axis, which is given by X
A
v ABy u sin α1 − u2 sin α2 h1
tan β = = 1
v ABx u1 cos α1 + u2 cos α2
x
For collision to take place,
h2 − h1
tan β = tan θ =
x
Ex.57 A particle A is projected with an initial velocity of 60 m/s. at an angle 30º to the horizontal. At
the same time a second particle B is projected in opposite direction with initial speed of 50 m/s
from a point at a distance of 100 m from A. If the particles collide in air, find (a) the angle of
projection α of particle B, (b) time when the collision takes place and (c) the distance of P from A,
where collision occurs. (g = 10 m/s2)
60m/s 50m/s
30º
A B
100m
Sol. (a) Taking x and y directions as
shown in figure. Y
Here, a = −gˆj , a = −gˆj
A B
4
(b) Now, | u AB |= u Ax – uBx = (30 3 + 50 cosα)m/s = 30 3 + 50 × 5 m / s = (30 3 + 40) m/s
AB 100
t= = or t = 1.09 s Ans.
| u AB | 30 3 + 40
(c) Distance of point P from A where collision takes place is
2 2
1 1
s= (u Ax t) 2 + u Ay t – gt 2 = . ) 2 + 30 × 109
( 30 3 × 109 . – × 10 × 109
. × 109
. or s = 62.64 m Ans.
2 2
5m/s
Ex.58 Two projectile are projected simultaneously from a point on the A
ground "O" and an elevated position "A" respectively as shonw in the
figure. If collision occurs at the point of return of two projectiles on
H
the horizontal surface, then find the height of "A" above the ground 10m/s
and the angle at which the projectile "O" at the ground
should be projected.
θ x
O C
Sol. There is no initial separation between two projectile is x-direction. For collision to occur, the relative
motion in x-direction should be zero. In other words, the component velocities in x-direction should be
equal to that two projetiles cover equal horizontal distance at any given time. Hence,
uOx = uAx
uA 5 1
⇒ u0cosθ = uA ⇒ cosθ = = = = cos60° ⇒ θ = 60°
uO 10 2
We should ensure that collision does occur at the point of return. It means that by the time projectiles
travel horizontal distances required, they should also cover vertical distances so that both projectile are
at "C" at the same time. In the nutshell, their times of flight should be equal.
For projectile from "O".
2uO sin θ
T=
g
For projectile from "A",
2H
T=
g
For projectile from "A",
2uo sin θ 2H
T= =
g g
Squaring both sides and putting values,
4u2O sin2 θ 4 × 102 sin2 60°
⇒ H= ⇒ H=
2g 2 × 10
2
3
H = 20 = 15m
2
We have deliberately worked out this problem taking advantage of the fact that projectiles are colliding
at the end of their flights and hence their times of flight should be equal. We can, however, proceed to
analyze in typical manner, using concept of relative velocity. The initial separation between two projectiles
in the vertical direction is "H". This separation is covered with the component of relative in vertical
direction.
3
⇒ vOAy = uOy – uAy = u0 sin60° – 0 = 10 × = 5 3m/s
2
Now, time of flight of projectile from ground is :
2uO sin θ 2x10x sin 60°
T= = = 3
g 10
Hence, the vertical displacement of projectile from "A" before collision is :
⇒ H = vOAy X T = 5 3 x 3 = 15 m/s
Ex.59 Two projectiles are projected simultaneously from two towers as shwon in figure. If the projectiles
collide in the air, then find the distance "s" between the towers.
10 m/s B
10 2 m / s
30m
A 45°
10m
Sol. We see here that projectiles are approaching both horizontally and vertically. Their movement in two
component directions should be synchronized so that they are at the same position at a particular given
time. For collision, the necessary requirement is that relative velocity and displacement should be in the
same direction.
It is given that collision does occur. It means that two projectiles should cover the displacement with
relative velocity in each of the component directions. Y 10 m/s B
In x-direction,
1 10 2 m / s
vABx = uAx – uBx = 10 2 cos 45° – (–10) = 10 2 + 10 = 20 m/s
2 30m
A 45°
If "t" is time after which collision occurs, then
⇒ s = vAy – uBy 10m
1
⇒ vABy = ucos45° – 0 = 10 2 × = 10m / s O S x
2
The initial vertical distance between points of projection is 30 – 10 = 20 m. This vertical distance is
covered with component of relative velocity in vertical direction. Hence, time taken to collide, "t", is :
20
⇒t= =2
10
Putting this value in the earlier equation for "s", we have :
⇒ s = 20t = 20x2 = 40 m
Now you can try all the questions related to relative motion.
9. Which of the following graphs cannot possibily 12. Give an example from your own experience in which
represent one dimensional motion of a particle? the velocity of an object is zero for just an instant of
x time, but its acceleration is not zero.
|v| l
Sol.
t t
t
Sol.
15. On a riverboat cruise, a plastic bottle is accidentally 18. A child is playing on the floor of a recreational
dropped overboard. A passenger on the boat estimate vehicle (RV) as it moves along the highway at a
that the boat pulls ahead of the bottle by 5 meters constant velocity. He has a toy cannon, which shoots
each second. Is it possible to conclude that the boat a marble at a fixed angle and speed with respect to
is moving at 5 m/s with respect to the shore? Account the floor. The connon can be aimed toward the front
for your answer. or the rear of the RV. Is the range towards the front
Sol. the same as, less than, or greater than the range
towards the rear? Answer this question (a) from the
child’s point of view and (b) from the point of view of
an observer standing still on the ground. Justify your
answers.
Sol.
9. The co-ordinates of a moving particle at a time t, 11. A body of mass 1 kg is acted upon by a force
are given by, x = 5 sin 10 t, y = 5 cos 10 t. The speed
F = 2 sin 3 πt i + 3 cos 3 πt j find its position at t = 1 sec
of the particle is -
(A) 25 (B) 50 (C) 10 (D) None if at t = 0 it is at rest at origin.
Sol. 3 3 2 2
(A) 2 , (B) 2 ,
3 π 9π 2 3π 3π 2
2 2
(C) , (D) none of these
3π 3π 2
Sol.
16. Balls are thrown vertically upward in such a way 18. The displacement-time graph of a moving particle
that the next ball is thrown when the previous one is is shown below. The instantaneous velocity of the
at the maximum height. If the maximum height is 5m, particle is negative at the point
the number of balls thrown per minute will be
x
(A) 40 (B) 50 (C) 60 (D) 120
Sol. D
E F
C
t
(A) C (B) D (C) E (D) F
Sol.
A C v(m/s)
60º
D
20
B 10
20. The displacement time graphs of two particles A 22. Acceleration versus velocity graph of a particle
and B are straight lines making angles of respectively
moving in a straight line starting form rest is as shown
30º and 60º with the time axis. If the velocity of A is
in figure. The corresponding velocity-time graph would
vA
vA and that of B is vB then the value of v is be -
B a
(A) 1/2 (B) 1 / 3 (C) 3 (D) 1/3
Sol.
v
v v
(A) (B)
t t
v v
v v
(A) (B)
t t
v v
(C) (D)
t t
Sol.
23. A man moves in x - y plane along the path shown. Question No. 25 to 27 (3 questions)
At what point is his average velocity vector in the
same direction as his instantaneous velocity vector. The x-t graph of a particle moving along a straight
The man starts from point P. line is shown in figure
y
C x parabola
PB D
A
x
(A) A (B) B (C) C (D) D 0 T 2T
Sol.
25. The v-t graph of the particle is correctly shown
by
v
v
T 2T
0 T 0
(A) 2T t (B) t
v v
0 T 2T 0 T 2T
(C) t (D) t
24. The acceleration of a particle which moves along
the positive x-axis varies with its position as shown.
If the velocity of the particle is 0.8 m/s at x = 0, the Sol.
velocity of the particle at x = 1.4 is (in m/s)
2
a (in m/s )
0.4
0.2
a v
0 0
(C) t (D) t
speed speed
(C) 0 0
T 2T t (D) T 2T t
Sol.
v(ms–1)
10
0
2 4 6 8 t(s)
–20
31. The maximum of displacement of the particle is 33. The correct displacement-time graph of the particle
(A) 33.3 m (B) 23.3 m (C) 18.3 (D) zero is shown as
x x
Sol.
(m) (m)
(A) (B)
0 2 4 6 8 t(s) 0 2 4 6 8 t(s)
x x
(m) (m)
(C) (D)
0 2 4 6 8 t(s) 0 2 4 6 8 t(s)
Sol.
(A) t (B)
t
V V
(C) (D)
t t
32. The total distance travelled by the particle is
(A) 66.6 m (B) 51.6 m (C) zero (D) 36.6 m
Sol. Sol.
35. Shown in the figure are the displacement time 37. A body A is thrown vertically upwards with such a
graph for two children going home from the school. velocity that it reaches a maximum height of h.
Which of the following statements about their relative Simultaneously another body B is dropped from height
motion is true after both of them started moving ? h. It strikes the ground and does not rebound. The
Their relative velocity: velocity of A relative to B v/s time graph is best
represented by : (upward direction is positive)
X
C1 VAB VAB
C2 (A) (B)
t t
O VAB VAB
t T
(A) first increases and then decreases t
(C) (D)
(B) first decreases and then increases
(C) is zero t
(D) is non zero constant Sol.
Sol.
100m
(A) 18 sec. (B) 16 sec. (C) 20 sec. (D) 17 sec.
Sol.
39. It takes one minute for a passenger standing on 42. A point mass is projected, making an acute angle
an escalator to reach the top. If the escalator does
with the horizontal. If angle between velocity v and
not move it takes him 3 minute to walk up. How long
will it take for the passenger to arrive at the top if he acceleration g is θ, then θ is given by
walks up the moving escalator? (A) 0º < θ < 90º (B) θ = 90º
(A) 30 sec (B) 45 sec (C) 40 sec (D) 35 sec (C) θ = 90º (D) 0º < θ < 180º
Sol. Sol.
Question No. 44 to 46
A projectile is thrown with a velocity of 50 ms–1 at an
angle of 53º with the horizontal
41. A ball is thrown upwards. It returns to ground
describing a parabolic path. Which of the following 44. Choose the incorrect statement
remains constant ? (A) It travels vertically with a velocity of 40 ms–1
(A) speed of the ball (B) It travels horizontally with a velocity of 30 ms–1
(B) kinetic energy of the ball (C) The minimum velocity of the projectile is 30 ms–1
(C) vertical component of velocity (D) None of these
(D) horizontal component of velocity. Sol.
Sol.
45. Determine the instants at which the projectile is 48. A ball is thrown from a point on ground at some
at the same height angle of projection. At the same time a bird starts
(A) t = 1s and t = 7s (B) t = 3s and t = 5s from a point directly above this point of projection at
(C) t = 2s and t = 6s (D) all the above a height h horizontally with speed u. Given that in its
Sol. flight ball just touches the bird at one point. Find the
distance on ground where ball strikes
h 2h 2h h
(A) 2u (B) u (C) 2u (D) u
g g g g
Sol.
50. A projectile is fired with a speed u at an angle θ Question No. 53 & 54 (2 questions)
with the horizontal. Its speed when its direction of At t = 0 a projectile is fired from a point O (taken as
motion makes an angle ‘α’ with the horizontal is - origin) on the ground with a speed of 50 m/s at an
(A) u secθ cosα (B) u secθ sinα angle of 53° with the horizontal. It just passes two
(C) u cosθ secα (D) u sinθ secα points A & B each at height 75 m above horizontal as
Sol. shown.
50m/s
A B
75m
53°
O
53. The horizontal separation between the points A
51. Two projectiles A and B are thrown with the same and B is -
speed such that A makes angle θ with the horizontal (A) 30 m (B) 60 m (C) 90 m (D) None
and B makes angle θ with the vertical, then - Sol.
(A) Both must have same time of flight
(B) Both must achieve same maximum height
(C) A must have more horizontal range than B
(D) Both may have same time of flight
Sol.
55. Particle is dropped from the height of 20 m from 58. One stone is projected horizontally from a 20 m
horizontal ground. There is wind blowing due to which high cliff with an initial speed of 10 ms–1. A second
horizontal acceleration of the particle becomes 6 ms–2. stone is simultaneously dropped from that cliff. Which
Find the horizontal displacement of the particle till it of the following is true ?
reaches ground. (A) Both strike the ground with the same velocity
(A) 6 m (B) 10 m (C) 12 m (D) 24 m (B) The ball with initial speed 10ms–1 reaches the
Sol. ground first
(C) Both the balls hit the ground at the same time
(D) One cannot say without knowing the height of
the building
Sol.
v
90° P
θ
61. On an inclined plane of inclination 30°, a ball is Q
thrown at an angle of 60° with the horizontal from the
(A) Tvsinθ (B) Tvcosθ (C) Tv secθ (D) Tv tanθ
foot of the incline with velocity of 10 3 ms–1. If g =
Sol.
10 ms–2, then the time in which ball with hit the inclined
plane is -
(A) 1.15 sec. (B) 6 sec
(C) 2 sec (D) 0.92 sec
Sol.
Sol. Sol.
70. A ship is travelling due east at 10 km/h. A ship 73. A swimmer’s speed in the direction of flow of river
heading 30º east of north is always due north from is 16 km h–1. Against the direction of flow of river, the
the first ship. The speed of the second ship in km/h is swimmer’s speed is 8 km h–1. Calculate the swimmer’s
(A) 20 2 (B) 20 3 / 2 (C) 20 (D) 20 / 2 speed in still water and the velocity of flow of the
Sol. river.
(A) 12 km/h, 4 km/h (B) 10 km/h, 3 km/h
(C) 10 km/h, 4 km/h (D) 12 km/h, 2 km/h
Sol.
v1
72. Four particles situated at the corners of a square
of side ‘a’ move at a constant speed v. Each particle
maintains a direction towards the next particle in
–1 1
succession. Calculate the time particles will take to (A) tan –1( 3) (B) tan
3
meet each other.
a a a 2a –1 1
(A) (B) (C) (D) (C) tan (D) None of these
v 2v 3v 3v 2
Sol. Sol.
Sol. Sol.
5. Let v and a denote the velocity and acceleration
respectively of a body in one-dimensional motion
7. Let v and a denote the velocity and acceleration
(A) | v| must decrease when a < 0 respectively of a body
(A) a can be non zero when v = 0
(B) Speed must increase when a > 0
(B) a must be zero when v = 0
(C) Speed will increase when both v and a are < 0 (C) a may be zero when v ≠ 0
(D) Speed will decrease when v < 0 and a > 0 (D) The direction of a must have some correlation
Sol. with the direction of v
Sol.
6. Which of the following statements are true for a 8. A bead is free to slide down a A
moving body? sm ooth wi re ti g ht l y st ret ched
(A) If its speed changes, its velocity must change between points A and B on a vertical θ
B R
and it must have some acceleration circle. If the bead starts from rest
(B) If its velocity changes, its speed must change at A, the highest point on the circle
and it must have some acceleration (A) its velocity v on arriving at B is proportional to
(C) If its velocity changes, its speed may or may not cosθ
change, and it must have some acceleration (B) its velocity v on arriving B is proportional to tanθ
(D) If its speed changes but direction of motion does (C) time to arrive at B is proportional to cosθ
not changes, its velocity may remain constant (D) time to arrive at B is independent of θ
+v0
v
T
O
t 2T
–v0
2 sec
(A) Car must move in circular path
(B) Acceleration of car is never zero 11. A block is thrown with a velocity of 2 ms–1 (relative
(C) Mean speed of the particle is π/4 m/s. to ground) on a belt, which is moving with velocity 4
(D) The car makes a turn once during its motion ms–1 in opposite direction of the initial velocity of block.
If the block stops slipping on the belt after 4 sec of
Sol.
the throwing then choose the correct statements(s)
(A) Displacement with respect to ground is zero after
2.66 sec and magnitude of displacement with respect
to ground is 12 m after 4 sec.
(B) Magnitude of displacement with respect to ground
in 4 sec is 4 m.
(C) Magnitude of displacement with respect to belt in
4 sec is 12 m.
(D) Displacement with respect to ground is zero in 8/
3 sec.
Sol.
t t X X
(A) y = 4h 1 − (B) y = 4h 1 −
T T R R
T T R R
(C) y = 4h 1 − (D) y = 4h 1 −
t t X X
Sol.
π
t= , & speed when t = π.
2
(b) Time when it crosses x-axis and y-axis
Sol. Sol.
60°
30°
x
2.5 sec
time in sec 2
o t 25 sec
Sol.
10
10 20 25 t
Sol.
gx2
21. The equation of a projectile is y = 3 x − . The
2
angle of projectile is ________ and initial velocity is
_______.
Sol.
25. A ball is projected on smooth inclined plane in 27. The horizontal range of a projectiles is R and the
direction perpendicular to line of greatest slope with maximum height attained by it is H. A strong wind now
velocity of 8m/s. Find it’s speed after 1 sec. begins to blow in the direction of motion of the
projectile, giving it a constant horizontal acceleration
= g/2. Under the same conditions of projection, find
8 m/s
the horizontal range of the projectile.
37º Sol.
Sol.
Sol.
A 37°
x
Sol.
20m/s
60º 45º
\\\\\\\\\\\\\\\\\\\\\\\\\\
10m
Sol.
60° 30°
20 m
Sol.
5. Two inclined planes OA and OB having inclination 6. A particle is thrown horizontally with relative velocity
(with horizontal) 30° and 60° respectively, intersect 10 m/s from an inclined plane, which is also moving
each other at O as shown in figure. A particle is with acceleration 10 m/s2 vertically upward. Find the
time after which it lands on the plane (g = 10 m/s2)
projected from point P with velocity u = 10 3 ms –1 along
a direction perpendicular to plane OA. If the particle 2
10 m/s
strikes plane OB perpendicularly at Q, calculate
A
u B 30°
Q
P Sol.
h
30° 60°
O
(a) velocity with which particle strikes the plane OB,
(b) time of flight,
(c) vertical height h of P from O,
(d) maximum height from O attained by the particle
and
(e) distance PQ
Sol.
P Q
45° x
Sol.
v=6m/s
Sol.
Exercise - V JEE-Problems
1. Two guns, situated at the top of a hill of height 10
m, fire one shot each with the same speed 5 3 m/s
at some interval of time. One gun fires horizontally
and other fires upwards at an angle of 60° with the
horizontal. The shots collide in air at a point P. Find
(a) the time interval between the firings, and
(b) the coordinates of the point P. Take origin of the
coordinates system at the foot of the hill right below
the muzzle and trajectories in X-Y plane.[JEE’ 1996]
Sol.
α
P Q
d
h h
(A) (B) d
v v
d d
(C) h (D) h
Sol.
u
O 3.0 m x
Sol.
45° 11 t(s)
O x
(a) The motion of the ball is observed from the frame (A) 110 m/s (B) 55 m/s (C) 550 m/s (D) 660 m/s
of trolley. Calculate the angle θ made by the velocity Sol.
vector of the ball with the x-axis in this frame.
(b) Find the speed of the ball with respect to the
4θ
surface, if φ = . [JEE 2002]
3
Sol.
v
v0
x0 x
a a
x
(A) (B)
x
a 12. A train is moving along a straight line with a con-
a
x x stant acceleration 'a'. A boy standing in the train
throws a ball forward with a speed of 10 m/s, at an
(C) (D) angle of 60° to the horizontal. The boy has to move
forward by 1.15 m inside the train to catch the ball
back at the initial height. The acceleration of the train
Sol. in m/s2 is [JEE’ 2011]
Sol.
ANSWER KEY
∆a = B − A = a 2 + a 2 + 2a 2 cos( π − dθ)
–a
= 2a2 (1 − cos θ) ⇒ 2a2 (1 − 1 + 2 sin 2dθ / 2) = 2a sin d θ/2
IIIrd Curve : Length of a moving body can not decrease with time
10 Ist Curve : A ball moving forward collides with surface rebounds and stops after IInd collision
IInd Curve : A ball repeatedly making inelastic collisions with floor.
IIIrd Curve : Collision of a ball with surface. {Surface has large velocity for short time}
11 (a) is incorrect car can not travel around track with constant velocity as direction is continuously
changing.
(b) correct
12 Ball at maximum height V = 0 for just an instant but acceleration due to gravity.
1
13 Vf = 2gH . Let balls meet after t sec. h1
1 1 H X
h1 = gt 2 and h2 = Vf t = gt 2
2 2 h2
H V0=Vf
h1 + h2 = H = Vf t H = 2gH t t = 2
2g
1 H H
∴ h1 = g = hence they will meet above half height of building.
2 2g 4
V2
∴ a= muzzle velocity is more for short barrl and S is also less hence acceleration will be more in that case.
2S
15 Hence we can not conclude that velocity of boat is 5 m/sec w.r.t. shore
VBottle = Vriver ; VB – VR = 5
16 Yes wrench will hit at the same place on the deck irrespective of that boat is at rest or moving because
when boat is at rest wrench will have zero horizontally velocity and when boat is moving both will have
same horizontal velocity.
17 Acceleration of the projectile remains constant throughout the journey = g
18 (a) In child point of view range will be same in both the cases.
(b) In ground frame of reference
VCT = VC – VT
VC = VCT + VT
For front range Vcannon = VC cos θ + VT Range will be more
For Rear range Vcannon = VC cos θ – VT Range will be less
d
19 d t= for tmin cos θ = 1 maximum Hence A will reach opposite end in least time
Vbr Vbr cos θ
15. D 16. C 17. B 18. C 19. C 20 D 21. C 22. D 23. C 24. B 25. B
26. D 27. C 28. B 29. C 30. A 31. A 32. A 33. C 34. A 35. D 36. D
37. C 38. C 39. B 40. B 41. D 42. D 43. B 44. A 45. D 46. A 47. B
48. C 49. B 50. C 51. D 52. B 53. B 54. A 55. C 56. C 57. C 58. C
59. A 60. A 61. C 62. C 63. D 64. B 65. C 66. D 67. B 68. D 69. D
70. C 71. A 72. A 73. A 74. A 75. B 76. B 77. C 78. B 79. B
10. A,B,C,D 11. B,C,D 12. A,C,D 13. A,B,C 14. C,D 15. D,C 16. A,B,C,D 17. A,B 18. A,B,C,D
19. A,C,D 20. B
1. (a) y2 + 8y + 12 = x ; (b) crosses x axis when t = 4 sec, crosses y axis when t = ± 2 sec.
1 3
4. (a) m / s , (b) m / s 2 , (c) 5. v = –30 i – 40 j, a = –16 i – 8 j 6. a/2b, a2/4b
3 2
2 38 v0 a vel π 2
7. (a) m , (b) m 8. 42 km/hr 9. 10. 3 cm/min 11. 25 m
3 3 – +
h
12. (a) 2.7 km; (b) 60 m/s; (c) 225 m and 2.25 km 13. 240 m 14. 15. 50 m
8g
16. 100 m, zero 17. 5 s 18. 36.2 sec. 19. 20 5 20. 20 sec 21. 60, 2 m/sec.
22. u = 50 ( 3 – 1) m/sec., H = 125 (– 3 + 2)m 23. (i) 1503.2 m (ii) 35.54 sec (iii) 3970.56 m
24. 100/3 m/s 25. 10 m/s 26. 75 m 27. R + 2H 28. 6 m/s 29. 20 × 2/3
–1 1 4
30. 10 m 31. 50 3 cm 32. tan–1 (1/2) 33. tan–1(3) 34. θ = tan ,
2 3
1
16.25 m, (e) 20 m 6. sec 7. 5 m/sec 8. 1 sec 9. 2 tan–1 (1/3) 10. θ = 37°, v = 6 m/s
3
a2 u 2 sin 2α u cos(α + θ)
1. (a) 1 sec, (b) ( (5 3 m, 5 m) 2. , tan –1 a 3. (a) , (b) v =
4b g cos θ cos θ