The document discusses uniformly accelerated motion, which is motion where the tangential acceleration is constant. It provides equations that relate the speed, distance, and acceleration of an object undergoing uniformly accelerated motion. Specifically, the speed and distance after a period of time can be determined based on the initial speed, acceleration, and time. The document also discusses applying these concepts to rigid bodies undergoing translational or rotational motion as well as the specific formulas for average acceleration and displacement.

1. 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.
Uniformly Accelerated Motion
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.
When a rigid body undergoes uniformly accelerated transla-tional 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
The Formulae
a = ( Vf - Vi ) / t
This is the formula for the average acceleration, which is our actual acceleration when
dealing with uniform acceleration. Average acceleration equals to the difference in
speed over time. When we have two speeds we can calculate the acceleration during the
time interval, t, by subtracting the initial speed from the final speed ( we get the change
in speed ) and then dividing by the time to get the acceleration.
Acceleration = increase in speed / time
d = 1/2 ( Vf + Vi ) × t
This formula is a favourite of mine. It stems from analyzing the graph of speed V.S. time
of an object's motion. Since speed = distance / time we can calculate the distance ( or
displacement ) an object by measuring the area under the line which represents speed.
The derivation of this formula is quite simple.
Using the Formulae
You now know two formulae that relate the initial speed, Vi, the final speed, Vf, the distance, d,
the acceleration, a and the time, t.
When you are solving a problem where you need to find one of these variables but you are

2. lacking another one, you should combine the two formulae to eliminate the unknown variable.
Here is a list of the combinations you can use when not given a variable:
Not Given Formula to eliminate that variable
( not needed )
d a = ( Vf - Vi ) / t
a d = 1/2 ( Vf + Vi ) × t
Vf 2d / t = at + 2Vi
Vi 2d / t = -at + 2Vf
t 2da = Vf² - Vi²
Problem 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?
Solution to Problem 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

3. Problem 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?
Solution to Problem 2:
a) The car has an initial velocity of 20 km/h, therefore the initial speed u = 20 km/h.
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
+ u t
where a is the acceleration (=8 m/s2
) and t is period of time between initial and final
positions and u is the initial velocity.
Since the time is given in seconds, we need to convert 20 km/h into m/s as follows:
u = 20 km/h =
20 * 1km
1 hour
1000 m
1 km
1 hour
3600 seconds
= 5.6 m/s
We now have
x = (1/2) (8) 102
+ 5.6*10 = 456 m
b) v = at + u = 8*10 + 5.6 = 85.6 m/s

4. Problem 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?
Solution to Problem 3:
a) The acceleration a is a measure if the rate of change of the velocity within a period of
time. Hence
u =
change in velocity
change in time
=
v - u
t
=
72 km/h - 0
11.5 seconds
We now convert 72 km/h into m/s
u = 72 km/h =
72 * 1km
1 hour
1000 m
1 km
1 hour
3600 seconds
= 20 m/s
We now calculate the acceleration a
a = (20 m/s) / (11.5 s) = 1.74 m/s2
(approximetd)
b) Two ways to find the position x:

5. 1) x = (1/2)(v + u) t or 2) x = (1/2) a t 2
+ u t
1) We first use: x = (1/2)(v + u) t = 0.5*(20 m/s + 0)*11.5 = 115 m
2) We now use: (1/2) a t2
+ u t = 0.5*1.74*(11.5) 2
+ 0*t = 115 m
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?
Solution to Problem 4:
a) We consider that the direction from ground up is the positive direction of the falling
object. We are given the initial (-20 m/s) and final velocities (-40 m/s); the minus sign
was added to take into account the fact that the falling object is moving in the negative
direction. We know the gravitational acceleration (g = - 9.8 m/s2
) acting on the falling
object and we are asked to find the height of the building. If we consider the position of
the object as being x (wth x = 0 on the ground), then we may use the equation relating

6. the initial and final velocities u and v, the acceleration a and the initial (x0 which the
height of the building) and final (x, on the ground) positions as follows:
v2
= u2
+ 2 a (x - x0)
(-40 m/s)2
= (-20 m/s)2
+ 2 (-9.8 m/s0) (0 - x0)
Solve the above for x0
x0 = 1200 / 19.6 = 61.2 m
b) x - x0 = (1/2)(u + v)t
-61.2 = 0.5(-20 - 40)t
t = 61.2 / 30 = 2.04 s
Problem 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?
Solution to Problem 5:
a) We are given the initial velocity u = 40 m/s, the final velocity v = 0 (train stops) and

7. 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.
Problem 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?

8. b) What distance was covered by the bicycle during the 10 seconds?
Solution to Problem 6:
a) In this problem the initial velocity u = 5 m/s and the final velocity v = 20 m/s. The
acceleration a of the bicycle is the rate of change of the velocity and is given as follows
a =
v - u
t
=
20 m/s - 5 m/s
10 seconds
= 1.5 m/s2
b) There are two ways to find the distance covered by the bicyle in t = 10 seconds.
1) x = (1/2)(v + u) t = 0.5 (20 + 5) 10 = 125 m
2) x = (1/2) a t2
+ u t = 0.5 * 1.5 * 100 + 5 * 10 = 125 m

9. Problem 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?
Solution to Problem 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)

10. Problem 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?
Solution to Problem 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.
Problem 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?
Solution to Problem 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

11. 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
Problem 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).
Solution to Problem 10:
a) 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

12. 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)
Problem 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?
Solution to Problem 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

13. = 4.9 t2
Solve for t: t = 1.42 seconds
b) u = 9.8 t = 9.8 * 1.24 = 14 m/s
Problem 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.
Solution to Problem 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

14. 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
total distance = 200 + 36,000 + 300 = 36,500 meters.
Example
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?

15. Here our knowledge of uniform acceleration is very useful. We are not given the
acceleration of the car as its stopping ( we assume the rate of acceleration is uniform )
and we need to find the distance.
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.
Lets do an example where we can combine the different formulae to get results.

16. Example
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?
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.