# SUVAT Equations

A LevelAQAEdexcelOCR

## SUVAT Equations

There are 5 SUVAT equations which relate 5 different variables of motion.

These 5 variables are:

• $\textcolor{red}{s} =$ Displacement
• $\textcolor{red}{u} =$ Initial velocity
• $\textcolor{red}{v} =$ Final velocity
• $\textcolor{red}{a} =$ Acceleration
• $\textcolor{red}{t} =$ Time taken

Questions will usually give you three variables, so you will have to figure out which equation to use.

Note: These equations can only be used when the acceleration is constant.

A LevelAQAEdexcelOCR

## Key Equations

• $\textcolor{red}{v} = u + \textcolor{blue}{a}\textcolor{purple}{t}$
• $\textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2$
• $\textcolor{limegreen}{s} = \dfrac{1}{2}(u + \textcolor{red}{v})\textcolor{purple}{t}$
• $\textcolor{red}v^2 = u^2 + 2\textcolor{blue}{a}\textcolor{limegreen}{s}$
• $\textcolor{limegreen}{s} = \textcolor{red}{v}\textcolor{purple}{t} - \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2$
A LevelAQAEdexcelOCR

## Paired Equations of Motion

We can use these equations to compare the motion of two (or more) separate particles.

For example, say we have two cars driving on the motorway. Car $1$ is in the slow lane, at a constant rate of $50\text{ mph}$. Car $2$ is overtaking from $\textcolor{limegreen}{160}\text{ m}$ behind, with an initial velocity of $60\text{ mph}$ and accelerates at a rate of $\textcolor{blue}{+2}\text{ mph per second}$.

Take $\textcolor{limegreen}{160}\text{ m} = \textcolor{limegreen}{0.1}\text{ miles}$.

Show that it will take approximately $\textcolor{purple}{36}\text{ seconds}$ for Car $2$ to overtake Car $1$.

Car 1: $u_1 = 50, \textcolor{blue}{a_1 = 0}, \textcolor{limegreen}{s_1 = x}\\$

As there is no acceleration, we can use the equation $\textcolor{limegreen}{s} = u\textcolor{purple}{t}\\$

$\textcolor{limegreen}{x} = 50\textcolor{purple}{t}\\$

Car 2: $u_2 = 60, \textcolor{blue}{a_2 = 2}, \textcolor{limegreen}{s_2 = x + 0.1}\\$

As we have the displacement, initial velocity, time taken and acceleration, we can use this SUVAT equation:

$\textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2\\$ $\textcolor{limegreen}{x + 0.1} = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\$

Solve these $2$ equations simultaneously:Â

$\textcolor{limegreen}{x} = 50\textcolor{purple}{t}\\$ $\textcolor{limegreen}{x + 0.1} = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\$

$50\textcolor{purple}{t} + 0.1 = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\$ $\textcolor{purple}{t}^2 + 10\textcolor{purple}{t} - 0.1 = 0\\$ $\textcolor{purple}{t} = \dfrac{-10 Â± \sqrt{100 + 0.4}}{2} = 0.00999...\text{ hrs and } -10.00999... \text{ hrs}$

Since time cannot be negative, the answer is

$0.00999...\text{ hrs} = \textcolor{purple}{35.96}\text{ seconds} â‰ˆ \textcolor{purple}{36}\text{ seconds}$

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## Note:

If a question states that an object is accelerating under gravity, take $a$ to be $g = 9.8\text{ ms}^{-2}$, unless stated otherwise.

A LevelAQAEdexcelOCR

## Example 1: Horizontal Motion

A car is driving at $20\text{ ms}^{-1}$. After $\textcolor{purple}{10}\text{ seconds}$, its speed is $\textcolor{red}{50}\text{ ms}^{-1}$.

How far does it travel in this time? What is its rate of acceleration?

[2 marks]

To work out the displacement, we need an equation using this and the known variables: displacement, initial velocity, time taken, and final velocity:

$\textcolor{limegreen}{s} = \dfrac{1}{2}(u + \textcolor{red}{v})\textcolor{purple}{t}\\$

$\textcolor{limegreen}{s} = \dfrac{1}{2}(20 + \textcolor{red}{50}) \times \textcolor{purple}{10} = \textcolor{limegreen}{350}\text{ m}\\$

And then acceleration is calculated by dividing the change in velocity by the time taken:

$\textcolor{blue}{a} = \dfrac{(\textcolor{red}{50} - 20)}{\textcolor{purple}{10}} = \textcolor{blue}{3}\text{ ms}^{-2}$

A LevelAQAEdexcelOCR

## Example 2: Vertical Motion

Danielle jumps from a diving board $\textcolor{limegreen}{10}\text{ m}$ above a pool. If her initial velocity (upwards) is $1\text{ ms}^{-1}$, how long does it take her to reach the water? What is her velocity at this point?

Assume $g = \textcolor{blue}{10}\text{ ms}^{-2}$.

[4 marks]

To find out time taken to reach the water, we need to use a SUVAT equation that uses time taken, displacement, initial velocity, and acceleration (as we can use gravity as acceleration):

$\textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2\\$ $\textcolor{limegreen}{10} = -\textcolor{purple}{t} + 5\textcolor{purple}{t}^2\\$ $5\textcolor{purple}{t}^2 - \textcolor{purple}{t} - \textcolor{limegreen}{10} = 0\\$ $\textcolor{purple}{t} = \textcolor{purple}{1.52}\text{ s (to } 2 \text{ dp)}\\$

And to work out her final velocity, we can use:

$\textcolor{red}v^2 = u^2 + 2\textcolor{blue}{a}\textcolor{limegreen}{s}\\$ $\textcolor{red}{v}^2 = (-1)^{-2} + (2 \times \textcolor{blue}{10} \times \textcolor{limegreen}{10})\\$ $\textcolor{red}{v} =\sqrt{201} = \textcolor{red}{14.18}\text{ ms}^{-1}$

A LevelAQAEdexcelOCR

## SUVAT Equations Example Questions

From the data provided, we have $s = 15\text{ m}$, $u = 0\text{ ms}^{-1}$, and $a = g$.

We need to find $v$.

Using the equation $v^2 = u^2 + 2as$, we have

$v^2 = 0^2 + (2 \times 9.8 \times 15)$

$v^2 = 2 \times 9.8 \times 15 = 294$

$v = \sqrt{294} = 17.15\text{ ms}^{-1}$

We know that $u = 480\text{ ms}^{-1}$, $v = 0\text{ ms}^{-1}$ and $t = 60\text{ s}$.

$v = u + at$ gives

$0 = 480 + 60a$

$a = \dfrac{-480}{60} = -8\text{ ms}^{-2}$

(A negative acceleration implies an object is slowing down or accelerating in the opposite direction)

For all three, $u = 0\text{ s and }s = 0.25\text{ mi}$.

James:

$a = 5.5\text{ mph/s} = 0.00152777...\text{ mi/s}\\$

$t = \sqrt{\dfrac{0.25 \times 2}{0.00152777}} = 18.09\text{ s}$

Jeremy:

$a = 7\text{ mph/s} = 0.0019444... \text{ mi/s}\\$

$t = \sqrt{\dfrac{0.25 \times 2}{0.0019444}} = 16.04\text{ s}$

Richard:

$a = 8\text{ mph/s} = 0.00222... \text{ mi/s}\\$

$t = \sqrt{\dfrac{0.25 \times 2}{0.00222}} = 15\text{ s}$

Removing $3 \text{ seconds}$ from James and $1 \text{ second}$ from Jeremy, we have

Richard – $15\text{ s}$

Jeremy – $15.04\text{ s}$

James – $15.09\text{ s}$

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