Constant acceleration formulae: SUVAT

​​In a nutshell

The formulae derived from the velocity-time graph: $$v=u+at$$ and $$s=\bigg(\dfrac{u+v}{2}\bigg)t$$, can be used to create three more formulae. These formulae are essential for solving problems about objects moving in a straight line with constant acceleration. There are five formulae in total.

Equations

Variable definitions

Deriving formulae

To derive the new formulae, you can eliminate variables from the previously derived formulae.

First eliminate $$t$$. 

To do this, first identify a formula to use:

​$$v=u+at$$

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Rearrange to make $$t$$ the subject:

​$$t=\dfrac{v-u}{a}$$

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Now bring in the other derived formula:

​$$s=\bigg(\dfrac{u+v}{2}\bigg)t$$

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Substitute $$t$$ with the rearranged expression:

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​$$s=\bigg(\dfrac{u+v}{2}\bigg)\bigg(\dfrac{v-u}{a}\bigg)=\dfrac{(u+v)(v-u)}{2a}$$

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Multiply both sides by $$2a$$:​

​$$2as=(u+v)(v-u)$$

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Expand the brackets:

​$$2as=v^2-u^2$$

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Rearrange to make $$v^2$$ the subject:​

​$$\boxed{v^2=u^2+2as}$$​​

Next, eliminate $$v$$.

Using a previously derived formula:

$$v=u+at$$

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Substitute this new expression for $$v$$ into the formula for displacement and simplify:

​$$s=(\dfrac{u+v}{2})t$$​​

​$$s=(\dfrac{u+u+at}{2})t$$​​

​$$s=(\dfrac{2u}{2}+\dfrac{at}{2})t$$​​

​$$s=(u+\dfrac{1}{2}at)t$$​​

​$$\boxed{s=ut+\dfrac{1}{2}at^2}$$​​

Finally, eliminate $$u$$.

Use a previously derived formula and make $$u$$ the subject:

​​$$v=u+at,\ therefore\ u=v-at$$​​

Substitute this into the new displacement equation and simplify:

$$s=ut+\dfrac{1}{2}at^2$$​​

​$$s=(v-at)t+\dfrac{1}{2}at^2$$​​

​$$s=vt-at^2+\dfrac{1}{2}at^2$$​​

​$$\boxed{s=vt-\dfrac{1}{2}at^2}$$​​

SUVAT

These five formulae make up the 'SUVAT' formulae. These formulae are known as constant acceleration formulae. The formulae are:

You should be able to derive and use each of these formulae for problems about objects moving in a straight line with constant acceleration.

Example

A particle moves in a straight line from A to B with a constant acceleration of $$4ms^{-2}$$. The velocity of the particle at point A is $$2ms^{-1}$$ and the velocity at B is $$20ms^{-1}$$.

a: Find the distance from A to B.

b: Find the time it takes the particle to go from A to B.

First, identify the variables which are given: $$u=2ms^{-1},\ v=20ms^{-1},\ a=4ms^{-2}$$.

a: The distance from A to B is also the displacement. The question does not give a value for $$t$$, therefore use a formula which does not include it:

$$v^2=u^2+2as$$​​

Substitute known values:

$$20^2 = 2^2+2(4)s$$​

$$400=4+8s$$​​

Rearrange to make $$s$$ the subject and solve:

$$8s=396$$

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$$s=\dfrac{396}{8}=49.5$$​​

The distance from $$A$$ to $$B$$ is $$\underline{49.5\ m}$$.

b: To avoid any potential errors in the working out for part a, use the equation which does not involve $$s$$:

$$v=u+at$$​​

Substitute known values:

$$20=2+4t$$​​

Rearrange to make $$t$$ the subject and solve:

$$4t=18$$

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$$t=\dfrac{18}{4} = 4.5$$​​

The particle takes $$\underline{ 4.5\ s}$$ to travel from $$A$$ to $$B$$.