: MOTION :

- Continuous change in position.

: TYPES OF MOTION :

A. Rectilinear Motion - Motion along a straight line.

A.1 Horizontal Rectilinear Motion, HRM
A.2 Vertical Rectilinear Motion, VRM (Freely Falling Body)
B. Projectile Motion - Motion of a body projected in space or plane.
C. Circular Motion - Motion along circular path.

: VELOCITY :

- A quantity that specifies how fast an object is moving and where it is going.
- Ratio of the displacement to the time consumed.

: AVERAGE VELOCITY : : INSTANTANEOUS VELOCITY :

- Represents the velocity of the - Represents the velocity of the object at a
object for the entire trip. point of the entire trip.
Vi - Velocity at initial point.
Displacement
v = Vf - Velocity at final point.
Time
: ACCELERATION :
- Change of velocity per unit time.

∆v
a =
∆t

Vi = 0 Vf = 5 m/s Vi = 5 m/s

a = + value

Vi = 5 m/s Vf = 0 a = + value
a = - value (angular acceleration)

Vf = 5 m/s
Vi = 5 m/s Vf = 5 m/s
a=0

VA = 0 VB = 5 m/s VC = 5 m/s VD = 0

a = + value a=0 a = - value

: UNIFORMLY ACCELERATED MOTION :
Time Velocity Acceleration
t1 = 0 V1 = 0 a = 5 m/s/s
}
}
t2 = 1 sec V2 = 5 m/s
t3 = 2 sec V3 = 10 m/s } a = 5 m/s/s a = 5 m/s2

t4 = 3 sec V4 = 15 m/s } a = 5 m/s/s

V1 = 0 V2 = 5 m/s V3 = 10 m/s V4 = 15 m/s

1 sec 1 sec 1 sec

: UNIFORMLY DECELERATED MOTION :

Time Velocity Acceleration
V1 = 15 m/s
t1 = 0
}
}
a = -5 m/s/s
t2 = 1 sec V2 = 10 m/s
t3 = 2 sec V3 = 5 m/s
} a = -5 m/s/s a = -5 m/s2

t4 = 3 sec V4 =0 } a = -5 m/s/s

V1 = 15 m/s V2 = 10 m/s V3 = 5 m/s V4 = 0

1 sec 1 sec 1 sec

: FORMULA FOR UNIFORMLY ACCELERATED/DECELERATED MOTION :

X
Consider: v = t

Vi Vf X=vt
(Vf + Vi)
X= (t)
2
X (Vi + at + Vi)
X= (t)
a 2
t X = Vit + ½ at2 2

∆V (Vf + Vi)
a= X= (t)
∆t 2
Vf - Vi
a= ti = 0 & tf = t (Vf + Vi) (Vf - Vi)
tf - ti X=
2 a
Vf - Vi
a= 2aX = Vf2 – Vi2
t
2aX + Vi2 = Vf2
Vf = Vi + at 1

Vf2 = Vi2 + 2aX 3

Problem: An airplane travels 800 m down the runway before taking off. It it starts
from rest, moves with constant acceleration, and becomes airborne in 20 s, what is
its speed when it takes off?

Vi = 0 Vf = ?

X = 800 m
a=?
t = 20 sec

Solve for “a” Substitute the value of “a”

from: into:
X = Vit + ½ at2 Vf = Vi + at

800 = (0)(20) + ½ (a)(20)2 Vf = 0 + (4)(20)

Vf = 80 m/s
1600 = 400(a)

a = 4 m/s2
Problem: A car moving at 20 m/s slows down at 1.5 m/s2 to a velocity of 10 m/s.
How far did the car go during the slowdown? How long did it last?

Vi = 20 m/s Vf = 10 m/s

It is a deceleration.
X=?
• •
a = - 1.5 m/s2
• •
t = ?
• •
Solve for “X” from: Solve for “t” from:

Vf2 = Vi2 + 2aX Vf = Vi + at

(10)2 = (20)2 + 2(-1.5)X 10 = 20 + (-1.5)t

100 = 400 - 3X 1.5t = 20 - 10

3X = 400 - 100 1.5t = 10

3X = 300 t = 6.67 sec
X = 100 m
Problem: A drag racer, starting from rest, speeds up for 402 m with an
acceleration of 17.0 m/s2. A parachute then opens, slowing the car down with an
acceleration of 6.10 m/s2. How fast is the racer moving 3.50 x 102 m after the
parachute opens?

VA = 0 VB = ? VC = ?

XAB = 402 m XBC = 350 m

aAB = +17.0 m/s2 aBC = -6.10 m/s2
tAB = ? tBC = ?

Consider Motion A to B: Consider Motion B to C:

Vf2 = Vi2 + 2aX Vf2 = Vi2 + 2aX

VB2 = VA2 + 2aABXAB VC2 = VB2 + 2aBCXBC

VB2 = (0)2 + 2(17)(402) VC2 = (116.91)2 + 2(-6.10)(350)

VB = 116.91 m/s VC = 96.94 m/s

Problem: A golf cart has an acceleration of 0.4 m/s2. What is its velocity after
it has covered 10 m starting from rest?

Solution:
From Formula 2:

Vi = 0 Vf = ? X = Vit + ½ at2

10 = (0)t + ½ (0.4)t2
X = 10 m
t = 7.07 s
a = 0.4 m/s2
t = ?
From Formula 1:

Vf = Vi + at
From Formula 3:
Vf2 = Vi2 + 2aX Vf = 0 + (0.4)(7.07)

Vf2 = (0)2 + 2(0.4)(10) Vf = 2.83 m/s

Vf = 2.83 m/s
: VERTICAL RECTILINEAR MOTION OR FREELY FALLING BODY :

: Assumptions :
i. The object is NOT acted upon by any other forces except its WEIGHT.
ii. Air resistance is neglected.
iii. Change in value of gravitational acceleration “g” is disregarded.
Consider: : SIGN CONVENTION :
A
• VA = 0
g = 9.8 m/s2
Going Down: The object is accelerating.
1 sec g = + V = + y = +
B
• VB = 9.8 m/s
Going Up: The object is decelerating.
1 sec g = 9.8 m/s2 g = - y = +
V = +
C
• VC = 19.6 m/s V=+ g = - (Always)
V = +,-
y =+
y = +,-
1 sec g = 9.8 m/s2 Ref. Line

y =-
V=-
D
• VD = 29.4 m/s
Recall: Now: Vi Vf
Ref. Line
Vi Vf

y y
g g
X
t t
a
t Ref. Line
Vf = Vi + at Vf Vi
X = Vit + ½ at2
Vf = Vi + gt Vf = Vi - gt
Vf2 = Vi2 + 2aX
y = Vit + ½ gt2 y = Vit - ½ gt2
: SEVEN POSSIBLE SET-UPS : Vf2 = Vi2 + 2gy Vf2 = Vi2 - 2gy
Vf = 0

Vf = -
Vf = + ymax = +
g=- y=+
y=+ g=-
g=- t=+
t=+ t=+ Vi = 0 Vi = -
y=- y=- y=-
Vi = + Vi = + Vi = + Vi = + Vf = - Vi = + g=- g=-
y=0 g=-
t=+ t=+ t=+
g=-
Vf = - Vf = - Vf = -
t=+
Problem: From the top of a cliff, a person uses a slingshot to fire a pebble straight
downward with an initial speed of 9.0 m/s. After 0.50 s, how far beneath the cliff-
top is the pebble?

Vi = - 9.0 m/s
Ref. Line y = Vit - ½ gt2
y= -?
y = (-9)(0.5) - ½(9.8)(0.5)2
g = 9.8 m/s2

t = 0.50 s y = -5.725 m

Vf = - ?
Problem: An apple is thrown vertically downward from a cliff 48 m high reaches the
ground in 2.0 s later. What was the apple’s initial velocity?

Solution:

Vi = ? y = Vit - ½ gt2
Ref. Line

y = - 48 m - 48 = Vi(2.0) - ½(9.8)(2.0)2

g = 9.8 m/s2
2Vi = ½(9.8)(2.0)2 - 48
t = 2.0 s
Vi = - 14.20 m/s
Vf = - ?
Problem: You are on the roof of the physics building, 46.0 m above the ground.
Your physics professor, who Is 1.80 m tall, is walking alongside the building at a
constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head,
where should the professor be when you release the egg?
Consider the Egg:
Solution: VA = 0
y = Vit – ½ gt2
yAB = - 44.2 m yAB = VAtAB – ½ gtAB2
g = 9.8 m/s2 -44.2 = 0 – ½ (9.8)tAB2
46.0 m
tAB = ? -44.2 = – 4.9tAB2
VB = ?
tAB = 3.0 s
1.8 m
Consider the Professor:
VC = 1.2 m/s XCD = ? VD = 1.2 m/s
aCD = 0 X = Vit + ½ at2
tCD = ?
XCD = VCtCD + ½ aCDtCD2

XCD = VCtCD

XCD = (1.2)(3.0)
XCD = 3.60 m