Equations of motion proof
In studying the motion of objects it is
often helpful to use an equation to work out the velocity, acceleration or the
distance travelled.
We use the following letters to represent certain
quantities:
| Distance travelled |
| s |
measured in metres (m) |
|
| Time taken |
| t |
measured in seconds (s) |
|
| Initial velocity |
| u |
measured in metres per second (ms-1) |
|
| Final velocity |
| v |
measured in metres per second (ms-1) |
|
| Acceleration |
| a |
measured in metres per second per second (ms-2) |
|
1. Non accelerated
motion – that is motion at a constant
velocity
The area
under the line of the velocity–time graph is the distance travelled by the
object in the time t.
For example u = 20m/s and t = 300 s
Distance (s) = ut = 20 x 300 = 6000 m
The equation for non accelerated
motion is:
Distance (s) = velocity (u or v) x time (t)
s = vt
2. Accelerated motion
Acceleration (a) = [change in velocity]/time = [v - u]/t
or
a = [v - u]/t
Another version is:
v = u + at
Distance travelled = area under the line =
ut + ˝ (v-u)t
But acceleration = (v-u)/t and so (v-u) = at
therefore:
Distance travelled (s) = ut + ˝ (v-u)t = ut + ˝ [at]t = ut +
˝ at
2
s = ut + ˝ at2
If the
object starts from rest u = 0 and so the equation becomes:
s = ˝ at2
Another useful equation
is:
v2 = u2 + 2as
This
equation can be proved as follows:
v = u + at therefore t = (v-u)/a but s = ut +
˝ at
2 and so
s = ut + ˝ a([v-u]/a)
2 therefore: 2s =
2u(v-u)/a + (v
2 – 2uv + u
2)/a
So: 2as = 2uv – 2u
2 + v
2
– 2uv + u
2 and so v
2 = u
2 + 2as
Equations summary
CONSTANT VELOCITY
1. s = vt
CONSTANT ACCELERATION
2. v = u + at
3. s = ut + ˝ at2
4. average velocity = [v + u]/t
5. v2 = u2 + 2as
USING
EQUATIONS
This section is designed to help you work out some of the problems
using the equations of motion.
Example problems
1. A horse accelerates steadily from rest at 4 ms-2 for 3s.
What is its final velocity and how far has it travelled?
(a) Initial velocity (u) = 0 Final velocity (v) = ?
Acceleration (a) = 4 ms-2 Time (t) = 3 s.
Known quantities u,a and t. Unknown v.
Correct equation is v = u + at
Final velocity (v) = 0 + 4x3 = 12 ms-1
(b) Correct equation is s = ut + ˝ at2
Distance (s) = 0x3 + 0.5x4x9 = 18 m
2. A car travelling at 20 ms-1 accelerates steadily at 5 ms-2 for a distance of 70 m. What is the final velocity of the car?
Known quantities u,a and s. Unknown v.
Correct equation is v2 = u2 + 2as
Final velocity2 (v2) = 30x30 + 2x5x70
Therefore v = 40 ms-1
If you need to use any of these equations to work
out problems the way to do it is this:
(a) write down what you are given, usually three
things
(b) look for the equation that contains these three things and the quantity that you are
trying to find
(c) put the numbers in the CORRECT equation and work it out
You will
need to know how to rearrange equations to make different quantities the subject of the
equation.
A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS USB
