### Acceleration

```Principles of Physics

motion along a straight line path, motion in
one dimension
 Which way are you headed?
 How far did you go?
 How fast are you going?
Acceleration
 Rate at which speed changes
 In other words, speeding up or slowing down
 Measured in meters per second2
 More commonly stated as “the object goes
from 0 to 40 m/s in 8 s”

Acceleration is calculated
2)
a
=
acceleration
(m/s
vf  vi
a
vf = final velocity (m/s)
t
vi = initial velocity (m/s)
v f  v i  at
t = time (s)

Acceleration is a vector quantity
 + when speeding up
 - when slowing down (deceleration)
Blue and green cars accelerate, the red car travels with a constant velocity.

At rest
Starting from rest → vi =0
Initially at rest

Comes to rest → vf = 0


1. A particle has a constant acceleration of 2.0 m/s2.
Calculate the time required for the particle to accelerate
from 8.0 m/s to 28 m/s.
Givens:
a = 2.0 m/s2
vf = 28 m/s
vi = 8.0 m/s
t=?
vf = vi +at
28 m/s = 8.0 m/s + 2.0 m/s2 (t)
28 m/s - 8.0 m/s = t
2.0 m/s2
10 s = t
2. A car travelling at 32 m/s comes to rest in 30 s.
Calculate the acceleration of the car.
Givens:
a=?
vf = 0 m/s
vi = 32 m/s
t = 30 s
vf  vi
a
t
0  32m / s
a
30s
a  1.07m / s 2
If you do not know the final velocity of an
object, but do know the displacement of the
object you can calculate the acceleration using:
d = vit + ½ at2
d = displacement (m)
vi = initial velocity (m/s)
t = time (s)
a = acceleration (m/s2)
3. A car starting from rest accelerates at 4 m/s2 for 6 s.
Calculate the car’s displacement.
Givens:
vi = 0 m/s
a = 4 m/s2
t= 6 s
d=?
d = vit + ½ at2
d = 0m/s (6s) + ½(4 m/s2 )(6s)2
d = 72 m
```