Constant acceleration: Deriving formulae

​​In a nutshell

Using the velocity-time graph, you can derive more formulae to work out the motion of an object with constant acceleration. 

Equations

Variable definitions

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Deriving formulae

You need to know how to derive two formulae from a velocity-time graph.

The acceleration of an object can be worked out using the gradient of a velocity-time graph: 

$$a=\dfrac{v-u}{t}$$​​

Rearrange this to make $$v$$ the subject:

​$$at=v-u$$

​​

$$\boxed{v = u+at}$$​​

The displacement of an object can be worked out by the area under the graph. You can work out the area using the formula for the area of a trapezium:

​$$Area\ of\ a\ trapezium= \bigg(\dfrac{a+b}{2}\bigg)h$$

In the velocity-time graph:​

$$a=u-0=u$$, $$b=v-0=v$$, $$h=t$$

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Substitute these values into the formula for the area of a trapezium:

​$$\boxed{s=\bigg(\dfrac{u+v}{2}\bigg)t}$$​​

These formulae can be rearranged to make each variable the subject, and then applied to objects in motion.

Example

An object starts at point A with a velocity of $$2\ ms^{-1}$$ and gets to point B after $$20$$​ seconds, reaching a velocity of $$10\ ms^{-1}$$. Work out:

a: The total displacement between points A and B.

b: The acceleration of the object between these points.

Start by identifying the known variables:

$$u=2\ ms^{-1}$$, $$v=10\ ms^{-1}$$, $$t=20\ s$$

a: Formula for displacement: 

$$s=\bigg(\dfrac{u+v}{2}\bigg)t$$

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Substitute the known variables:

$$s=\Big( \dfrac{2+10}{2}\Big)20$$​​

$$s = \bigg(\dfrac{12}{2}\bigg)20 = 6\times 20 = 120$$​​

The displacement of the object is $$\underline{ 120 \ m}$$.

b: Formula for acceleration:

$$a=\dfrac{v-u}{t}$$

Substitute the known variables:

$$a=\dfrac{10-2}{20}=\dfrac{8}{20}=\dfrac{4}{10}$$​​

The acceleration of the object is $$\underline{0.4 \ m\cdot s^{-2}}$$.