# The SUVAT Equations and their derivation

The SUVAT Equations describe motion in a given direction when **ACCELERATION IS CONSTANT**. The SUVAT Equations that are given in the Formula Booklet are:

- $\hspace{10pt}V=U+AT$
- $\hspace{10pt}S=\left(\frac{U+V}{2}\right)T$
- $\hspace{10pt}V^2=U^2+2AS$
- ……. SEE ALL SUVAT EQUATIONS

Variable | Description | SI unit |
---|---|---|

S | displacement | $m$ |

U | initial velocity | $m/s$ |

V | final velocity | $m/s$ |

A | acceleration | $m/s^2$ |

T | total time | $s$ |

## Derivation of the SUVAT Equations

In general, for motion in a straight line with constant acceleration:

$\text{acceleration} = \frac{\text{change in velocity}}{\text{change in time}}\hspace{3pt}\text{or}\hspace{3pt}A=\frac{V-U}{T}$

where V is the final velocity, U is the initial velocity and T is the total time taken. Rearranging gives the equation in an alternative form:

$V=U+AT$

This equation is one of the **SUVAT **equations. They are named so since they involve displacement (S), initial velocity (U), final velocity (V), acceleration (A) and time (T) for motion in a straight line with constant acceleration. Note that in order to use the SUVAT equations, we must define a reference point with regards to the displacement. We must also specify the direction of positive and negative speed. Speed cannot be negative but acceleration can. If an object is slowing down rather than speeding up, acceleration is negative.

The second SUVAT equation $S=\frac{1}{2}(U+V)T$ comes from the fact that acceleration is constant. In this case, $\frac{1}{2}(U+V)$ is the average speed throughout the duration of travel. Multiplying this by T will give the total distance. This is because distance is speed multiplied by time when acceleration is constant.

The first two SUVAT equations can be used to derive the remaining SUVAT equations:

- SUVAT Equation 1 can be rearranged to make T the subject so that $T=\frac{V-U}{A}$ which can be substituted into equation 2:

$S=\left(\frac{U+V}{2}\right)\left(\frac{V-U}{A}\right)$ and rearranged gives $V^2=U^2+2AS$ - Substitute the expression for V in SUVAT Equation 1 directly into SUVAT Equation 2:

$\begin{array}{l}S&=&\left(\frac{U+U+AT}{2}\right)T\\&=&UT+\frac{1}{2}AT^2\end{array}$ - SUVAT Equation 1 can be rearranged to make U the subject so that $U=V-AT$. Substitute this into equation 4 to give

$\begin{array}{l}S&=&\left(V-AT\right)T+\frac{1}{2}AT^2\\&=&VT-\frac{1}{2}AT^2\end{array}$

See the Examples for use of the SUVAT equations and an example derivation question. When attempting examples for yourself, make sure that the dimensions are consistent. In other words, you should ensure that you are using the same SI unit for all measurements. Accordingly, this may require a conversion. See Motion under Gravity for more examples using the SUVAT equations.

## Examples

### Example 1

A ball starts rolling down a hill from rest with a constant acceleration of 6.7 m/s/s. Find the velocity of the ball after 4 seconds. How far has the ball travelled in that time?

Firstly, SUVAT equation 1 will determine the velocity after 4 seconds. Clearly, the initial velocity is 0, the acceleration is 6.7 and the time is 4. Hence, the final velocity is $V=0+6.7\times 4=26.8$ m/s.

Secondly, the displacement or how far the ball travelled will be given by SUVAT equation 2. Again the initial velocity is 0 and the time is 4. This time final velocity is 26.8. It follows that the distance travelled is $S=\left(\frac{0+26.8}{2}\right)\times 4=53.6$ metres.

### Example 2

A cyclist is in a race and 100 metres from the finish he decides to accelerate his speed. The cyclist maintains a constant acceleration of 0.4m/s/s. If the cyclist crosses the finish line with a speed of 17m/s, how fast was he cycling when he started to accelerate?

Use the SUVAT equation that involves S, A, V and U to answer this question. Evidently, this is SUVAT equation 3 where we wish to calculate U. Rearranging SUVAT equation 3 gives $U=\sqrt{V^2-2AS}$. Substituting S=100, A=0.4 and V=17, we have that the cyclist must have started accelerating when his speed was

$U=\sqrt{209}=14.5$ m/s to 1 decimal place.

### Example 3

A particle is moving along a line parallel with the x-axis. The particle is observed at the origin with a speed of 12 micrometres per second. Point A is 160 micrometres from the origin. If the particle decelerates at a constant rate of 0.4 micrometres per second per second, find the times that the particle passes point A.

A micrometre ($\mu m$) is one millionth of a metre. Since all measurements are in micrometres, however, we do not need to make the conversion to metres.

We have displacement, acceleration (or deceleration in this case), initial speed and we want to find time. Consequently, we should use SUVAT equation 4. Substituting S=160, U=12 and A=-0.4 we have $160=12T-\frac{1}{2}\times 0.4\times T^2$. Some rearranging and simplifying gives $T^2-60T+800=0$. This quadratic can be solved by factorising: $(T-20)(T-40)=0$. Hence, the particle passes the point A after 20 and 40 seconds. This suggests that the particle, travelling in the positive x-direction, decelerates for a sufficiently long time to go back past A this time travelling in the negative x-direction.

### Example 4

Consider the expression for final velocity $V=U+AT$ where U is initial velocity, A is constant acceleration and T is time. Given also that displacement S is given by $S=\frac{1}{2}(U+V)T$, show that $V^2=U^2+2AS$.

Rearrange $V=U+AT$ to make T the subject:

$V=U+AT\hspace{3pt} \Rightarrow\hspace{3pt} V-U=AT\hspace{3pt}\Rightarrow\hspace{3pt}\left(\frac{V-U}{A}\right)=T$

Substituting this expression for T into the given expression for S gives:

$S=\frac{1}{2}(U+V)\times\left(\frac{V-U}{A}\right)=\frac{1}{2A}(U+V)(V-U)$

Multiplying both sides by 2A and expanding the brackets gives:

$2AS=U^2-V^2\hspace{4pt}\Rightarrow\hspace{4pt}V^2=U^2-2AS$

## Questions by Topic

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