1. Equations of Motion

2. Vectors – size (magnitude) and direction
Add head to tail for relative velocity problems (eg boat on river)
Subtract by adding the opposite for change problems (final – initial!!!!)
Components – vertical and horizontal (projectiles)

3. Our basic formulae
v = Δd Δt
a = Δv
Used for uniform motion.

4. Graphs of Motion

5. The most useful graph is
The velocity time graph because from this we can get the
velocity (read graph)
acceleration (the gradient)
distance travelled (the area under the graph)

6. To solve this problem
A dragster travelling at a constant speed accelerates uniformly from 56ms-1 for 8s to reach a new speed of 85ms-1. What distance did it do and what was the acceleration?

7. Equations of motion
Using a general velocity- time graph we can look at any situation with constant (or zero) acceleration and get some useful equations

8. Using the gradient (acceleration)

9. Using the area under the graph

10. Using tricky maths we get a 3rd equation
By substituting for t from vf = vi + at
into d = vit + 1/2at2 and rearranging, it can be shown that
These equations are called the kinematic equations of motion

11. Solving problems using the kinematic eqations of motion
An object is decelerating at 0.24ms-2 from an initial velocity of 22ms-1. Find the velocity of the object after half a minute

12. Steps to solve problem
List what you know (making sure units are SI units)
Decide what you want to find out
Select the best equation
Rearrange, substitute and solve
a = -0.24ms-2
(- because deceleration)
vi = 22ms-1
t = 0.5min = 30s
vf = ?
vf = vi + at
= 22 – 0.24 x 30
= 14.8 ms-1

13. Try one for your self
A car accelerates from rest at a constant rate of 8ms-2.
How fast is it going when it is 100m from its starting position?
vi = 0 ms-1
a = 8 ms-2
d = 100m
vf = ?
vf2 = vi2 + 2ad
= x 8 x 100
= 1600
vf = 40ms-1

14. A mathematical note
In physics, when multiplying and dividing your answer should be no more accurate than the least accuarate factor (rounding to sig figs).
3601 (4sf) x (5sf) = (4sf)
3601 (4sf) x (2sf) = (2sf)
100 could be 1, 2 or 3 sf. Unless told otherwise, assume it to be the most accurate (ie 3 sf)