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KINEMATICS

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KINEMATICS

Kinematics is the study of the motion of objects without concern for the forces causing the motion.

v = the magnitude of the velocity of the object (meters per second, m/s)
v1 = the magnitude of the initial velocity (meters per second, m/s) (in some texts this is vi or v 0)
v2 = the magnitude of the final velocity (meters per second, m/s) (in some texts this is v f)
a = the magnitude of the acceleration (in meters per second squared, m/s2)
s = the displacement vector, the magnitude of the displacement is the distance,
s = │s│ = d (vectors are indicated in bold; the same symbol not in bold represents the magnitude of the
vector)
Δ indicates change, for example Δv = (v2 –v1)
t = time
t1 = the initial time
t2 = the final time

To derive the big 4 Equation we analyze a or acceleration as slope of the line and recognize as a linear
acceleration y=mx+b the 1st equation will arise

To get the next equation, derive an expression for the displacement of the object during the time
interval, Δt. The displacement for an object traveling at a constant velocity can be found by multiplying
the object’s velocity by the time the object travels at that velocity.
The displacement of the object in Fig. 2 is equal to v1Δt. The product, v1Δt, is also equal to the area
A1. Similarly, the total displacement of the object in Fig. 1, which is moving at a changing velocity, can
be found by calculating the area under the line during the time interval, Δt. For some objects this
calculation can be a little tricky, but for the object depicted in Fig. 1, calculating the area under the
line simply means calculating the area of the rectangle A1 and the triangle A2 and adding the values.
Area A1 is a rectangle. The length, l, is v1. The width, w, is t2 – t1, (Δt).
Area of rectangle A1 = l x w

Combining all of those equation.. Equation two arises

Equation 2 and !1 combined Equation 3 arises


Multiply the left side of Equation 2 by the left side of Equation 1, and multiply the right side of Equation
2 by the right side of Equation 1. Doing this means multiplying both sides by acceleration, but this will
allow Δt to cancel on the right side of the equation.

Thus the Big four equation of kinematics

In this case these are called the equation of Linear motion or Rectillinear Motionis a one-
dimensional motion along a straight line, and can therefore be described mathematically using only one
spatial dimension. The linear motion can be of two types: uniform linear motion with constant velocity
or zero acceleration; non uniform linear motion with variable velocity or non-zero acceleration. The
motion of a particle (a point-like object) along a line can be described by its position x, which varies with
t (time).
1. On a certain stretch of track, trains run at 60 mph (96.56 kph). How far back of a stopped
train should be a warning torpedo be placed to signal an oncoming train? Assume that the
brakes are applied at once and retard the train at the uniform rate of 2 ft/sec2 (0.61 m/s2).
s=1936 ft s=589.7 m
2. A ball is dropped from the top of a tower 80 ft (24.38 m) high at the same instant that a
second ball is thrown upward from the ground with an initial velocity of 40 ft/sec (12.19
m/s). When and where do they pass, and with what relative velocity? vr=40 ft/sec Time
and height: 2 seconds at 64.4 ft
3. Particles A and B are elevated 12 meters high from a given reference base. Particle A is
projected down an incline of length 20 meters at the same time particle B is let to freely fall
vertically. Find the velocity of projection of particle A if both particles strikes the base at the
same time.

4. A stone is thrown vertically up from the ground with a velocity of 300 ft per sec (91.44 m/s).
How long must one wait before dropping a second stone from the top of a 600-ft (182.88-m)
tower if the two stones are to pass each other 200 ft (60.96 m) from the top of the tower?
t2=13.67 sec

5. A stone is thrown vertically upward from the ground with a velocity of 48.3 ft per sec
(14.72 m per sec). One second later another stone is thrown vertically upward with a
velocity of 96.6 ft per sec (29.44 m per sec). How far above the ground will the stones
be at the same level?

6. The motion of a particle is given by the equation s=2t4−16t3+2t2s=2t4−16t3+2t2


where ss is in meter and tt in seconds. Compute the values of vv and aa when
t=2 sect=2 sec.

7. A stone is thrown vertically up from the ground with a velocity of 300 ft per sec
(91.44 m/s). How long must one wait before dropping a second stone from the top of
a 600-ft (182.88-m) tower if the two stones are to pass each other 200 ft (60.96 m)
from the top of the tower?
Projectile motion(Curvillinear Motion)

Projectile Definition

When an object is inflight after being projected or thrown then that object is called a projectile and this
motion is called Projectile Motion. Example, the motions of a cricket ball, baseball.

Projectile motion- the motion of a projected object in flight is known as projectile motion which is a
result of 2 separate simultaneously occurring components of motions. One component is along a
horizontal direction without any acceleration and the other along the vertical direction with constant
acceleration due to the force of gravity.

Equation of motion
RANGE AND HEIGHT

1. A projectile is fired up the inclined plane at an initial velocity of 15 m/s. The plane is making an
angle of 30° from the horizontal. If the projectile was fired at 30° from the incline, compute the
maximum height z measured perpendicular to the incline that is reached by the projectile.
Neglect air resistance. z=3.31 m

2. A bullet is fired at an initial velocity of 150 m/s and an angle of 56° at the top of a 120 m tall
building. Neglecting air resistance, determine the following:

a) The maximum height above the level ground that can be reached by the bullet. Hmax=908.19 m

b) The time for the bullet to hit the ground. t=26.284 sec

c) The velocity with which the bullet will hit the ground. vG=157.656 m/sec
3. An aircraft is moving horizontally with a speed of 50 m/s. At the height of 2 km, an object is
dropped from the aircraft. Acceleration due to gravity = 10 m/s2, what is the time interval
before the object hits the ground.

4. A kicked football leaves the ground at an angle θ = 45o with the horizontal has an initial speed of
25 m/s. Determine the distance of X. Acceleration due to gravity is 10 m/s2.

5. In the given picture you see the motion path of cannonball. Find the maximum height it can
reach, horizontal distance it covers and total time from the given information. (The angle
between cannonball and horizontal is 53º )

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