1. UNIFORMLY
ACCELERATED
MOTION

2. UNIFORMLY ACCELERATED MOTION
 Uniformly Accelerated Motion is the
motion of a point such that its
tangential acceleration w is constant.
In the case of uniformly accelerated
rectilinear motion, the point's
acceleration w is constant. The speed
? of the point t sec after uniform
acceleration begins and the point's
distance s from its initial position are
determined for uniformly accelerated
motion by the equations.

3.  the motion of a point such that its
tangential acceleration wτ is constant; in
the case of uniformly accelerated
rectilinear motion, the point’s
acceleration w is constant. The speed ν
of the point t sec after uniform
acceleration begins and the point’s
distance s from its initial position—s
being measured along the point’s path—
are determined for uniformly accelerated
motion by the equations
 ν = ν0 + wτt s = v0t + wτt2/2
 where ν0 is the initial speed of the point.
When ν and wT are of the same
sign, acceleration occurs; when they are
of opposite sign, deceleration occurs.

4.  When a rigid body undergoes
uniformly accelerated translational
motion, the above definitions apply to
each point of the body. A body may
also undergo uniformly accelerated
rotation about a fixed axis; in this
case, the body’s angular acceleration;
ε is constant, and the angular speed ω
and angular displacement ɸ of the
body are
ω = ω0 + ∊ t ɸ = ω0t t + ∊t2/2

5. EXAMPLE
PROBLEMS

6. PROBLEM NO.1
 From rest, a car accelerated at 8 m/s2
for 10 seconds.
a) What is the position of the car at the
end of the 10 seconds?
b) What is the velocity of the car at the
end of the 10 seconds?

7. SOLUTION TO PROBLEM NO.1
a) The car starts from rest therefore the initial speed u = 0.
Nothing is said about the initial position and we
therefore assume it is equal to 0. Hence the position x
is given by the equation
x = (1/2) a t 2
where a is the acceleration (=8 m/s2) and t is the period
of time between initial and final positions
x = (1/2)8 (10)2 = 400 m
b) The velocity v of the car at the end of the 10 seconds
is given by
v = a t = 8 * 10 = 80 m/s

8. PROBLEM NO. 2
 With an initial velocity of 20 km/h, a
car accelerated at 8 m/s2 for 10
seconds.
a) What is the position of the car at the
end of the 10 seconds?
b) What is the velocity of the car at the
end of the 10 seconds?

9. SOLUTION TO PROBLEM NO.
2

10. PROBLEM NO. 3
 A car accelerates uniformly from 0 to
72 km/h in 11.5 seconds.
a) What is the acceleration of the car
in m/s2?
b) What is the position of the car by
the time it reaches the velocity of 72
km/h?

11. SOLUTION TO PROBLEM NO.
3

12. PROBLEM NO. 4
 Problem 4: An object is thrown straight
down from the top of a building at a
speed of 20 m/s. It hits the ground
with a speed of 40 m/s.
a) How high is the building?
b) How long was the object in the air?

13. SOLUTION TO PROBLEM NO.
4

14. PROBLEM NO. 5
 A train brakes from 40 m/s to a stop
over a distance of 100 m.
a) What is the acceleration of the
train?
b) How much time does it take the
train to stop?

15. SOLUTION TO PROBLEM NO.
5a) We are given the initial velocity u = 40 m/s, the final velocity v = 0 (train
stops) and the distance. Hence the formula that relates these 3 quantities
and the acceleration is given by
v2 = u2 + 2 a x
02 = 402 + 2 a (100)
Solve for the acceleration a
a = -1600 / 200 = - 8 m/s2
b) There two ways to find the time:
1) Use: x = (1/2)(v + u) t
100 = 0.5(0 + 40) t
Solve for t: t = 5 seconds.
2) Use x = (1/2) a t2 + ut
100 = 0.5 ( - 8) t2 + 40t
4 t2 - 40 t + 100 = 0
4 (t2 - 10 t + 25) = 0
4(t - 5)2 = 0
t = 5 seconds.

16. PROBLEM NO. 6
 A boy on a bicycle increases his
velocity from 5 m/s to 20 m/s in 10
seconds.
a) What is the acceleration of the
bicycle?
b) What distance was covered by the
bicycle during the 10 seconds?

17. SOLUTION TO PROBLEM NO.
6

18. PROBLEM NO. 7
 a) How long does it take an airplane to
take off if it needs to reach a speed on
the ground of 350 km/h over a
distance of 600 meters (assume the
plane starts from rest)?
b) What is the acceleration of the
airplane over the 600 meters?

19. SOLUTION TO PROBLEM NO.
7
a) In this problem the initial velocity u = 0 (assumed because it is not
given) , the final velocity v = 350 km/h and the distance x = 600
meters = 0.6 km
The relationship between the give quantities is:
x = (1/2)(v + u) t
0.6 = 0.5 (350 + 0) t
Solve for t
t = (0.6 / 175) hours = 12.3 seconds
b) The acceleration a of the airplane is given by
a = (v - u) / t = 350 km/h / 12.3 s
Convert 350 km/h into m/s
350 km/h = 350,000 m / 3,600 s = 97.2 m/s
a = 97.2 m/s / 12.3 s = 8 m/s2 (to the nearest unit)

20. PROBLEM NO. 8
 Starting from a distance of 20 meters
to the left of the origin and at a velocity
of 10 m/s, an object accelerates to the
right of the origin for 5 seconds at 4
m/s2. What is the position of the object
at the end of the 5 seconds of
acceleration?

21. SOLUTION TO PROBLEM NO.
8
 a) In this problem, we may consider that the
direction of the object is the positive direction
and the initial position x0 = -20 meters (to the
left of the origin), the initial velocity u = 10 m/s,
the acceleration a = 4 m/s2 and the time is t =
5 seconds. The position is given by
x = (1/2) a t2 + u t + x0
= 0.5 * 4 * (5)2 + 10 * 5 - 20 = 80 meters to the
right of the origin.

22. PROBLEM NO. 9
 What is the smallest distance, in
meters, needed for an airplane
touching the runway with a velocity of
360 km/h and an acceleration of -10
m/s2 to come to rest?

23. SOLUTION TO PROBLEM NO.
9
a) In this problem the initial velocity u = 360
km/h, the final velocity v = 0 (rest) and the
acceleration a = -10 m/s2. The distance x can
be calculated using the formula
v2 = u2 + 2 a x
Convert 360 km/h into m/s: 360 km/h = (360
000 m) /(3600 s) = 100 m/s
x = ( v2 - u2 ) / (2 a) = (0 - 10,000) / (-20) = 500
meters

24. PROBLEM NO. 10
 To approximate the height of a water
well, Martha and John drop a heavy
rock into the well. 8 seconds after the
rock is dropped, they hear a splash
caused by the impact of the rock on
the water. What is the height of the
well. (Speed of sound in air is 340
m/s).

25. SOLUTION TO PROBLEM NO.
10a) In this problem we have:
1) a rock was dropped down the well and is uniformly accelerated downward due
to gravity. If h is the height of the well and t is the time taken by the rock to reach
the bottom of the well, then we have
h = (1/2)(9.8) t 2
2) After the splash, the sound travels up the well at a constant speed of 340 m/s.
Again the same height h of the well is given by
h = 340 *(8 - t) : 8 - t is the time taken for the sount to travel from bottom to top
where the sound is heard.
The above equations give:
(1/2)(9.8) t2 = 340 *(8 - t)
4.9 t2 + 340 t - 2720 = 0
Solve for t, two solutions:
t = 7.24 s and the second solution is negative and is not valid.
The height h of the well is calculated using one of the above equations:
h = 340 *(8 - t) = 340 *(8 - 7.24) = 257 meters (approximated to the the nearest
meter)

26. PROBLEM NO. 11
 A rock is thrown straight up and
reaches a height of 10 m.
a) How long was the rock in the air?
b) What is the initial velocity of the
rock?

27. SOLUTION TO PROBLEM NO.
11
 a) In this problem the rock has an initial velocity u.
When the rock reaches a height of 10 m, it returns
down to earth and the the velocity v = 0 when x = 10
meters. Hence
v = -9.8 t + u
0 = -9.8 t + u
u = 9.8 t
x = (1/2)(u + v) t
10 = 0.5 (9.8 t + 0) t
= 4.9 t2
Solve for t: t = 1.42 seconds
b) u = 9.8 t = 9.8 * 1.24 = 14 m/s

28. PROBLEM NO. 12
 A car accelerates from rest at 1.0 m/s2
for 20.0 seconds along a straight road
. It then moves at a constant speed for
half an hour. It then decelerates
uniformly to a stop in 30.0 s. Find the
total distance covered by the car.

29. SOLUTION TO PROBLEM NO.
12
a) The car goes through 3 stages:
stage 1: acceleration a = 1, initial velocity = 0, t = 20 s.
Hence the distance x is given by
x = (1/2) a t2 = (1/2) (1) 202 = 200 meters
stage 2: constant speed v is the speed at the end of stage
1.
v = a t = 1 * 20 = 20 m/s
x = v t = 20 m/s * (1/2 hour) = 20 m/s * 1800 s = 36,000
meters
stage 3: deceleration to a stop, hence u = 20 m/s and v = 0
(stop)
x = (1/2)(u + v) t = (1/2)(20 + 0) 30 = 300 meters

30. PROBLEM NO. 13
 A car was travelling at a speed of
70km/h, the driver saw a rabbit on the
road and slammed on the breaks.
After 6.0 seconds the car came to a
halt, how far did the car travel from the
point where the brakes were first
pressed to the point where the car
stopped?

31. SOLUTION TO PROBLEM NO.
13
 We are given:
Vi = 70km/h = 19.4 m/s
Vf = 0km/h
t = 6s
d = ?
Our formula for distance is d = 1/2 ( Vf + Vi ) × t
d = 1/2 ( 19.4 + 0 ) × 6
d = 58.332
 The car stopped after 58 metres.
This car has very bad brakes, as 58 metres is a very
long distance to come to a halt.

32. PROBLEM NO. 14
 Bill jogs at 6.0km/h, he then decides to
accelerate into a light run.
Bill accelerates at 0.030km/s² as he
runs through a distance of 40m
What is Bill's final speed?

33. SOLUTION TO PROBLEM NO.
14
 First we convert our given information into a uniform set of
magnitudes, metres and seconds. We are given:
Vi = 6.0km/h = 1.6m/s
Vf = ?
d = 40m
a = 0.003km/s² = 0.30m/s²
 Lets rearrange our uniform acceleration equations to
eliminate the time, which we are not given.
2da = Vf² - Vi²
Vf² = 2da + Vi²
Vf² = 2(40)(0.3) + 1.6²
Vf² = 26.6
Vf = 5.2m/s
Vf = 18.5km/h
 After accelerating Bill ends up running at 18 km/h.

34. PROBLEM NO. 15
 In performing a dismount off the high
bar (3.048 meters above the ground),
a gymnast's center of gravity reaches
a height above the bar, of 2.5 m. Find:
a. What was the gymnast's takeoff
velocity?
b. What was the velocity of the
gymnast at landing?

35. SOLUTION TO PROBLEM NO.
15