# SUVAT Equations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## 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.

## 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 Level

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

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

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

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

$50\textcolor{purple}{t} + 0.1 = 60\textcolor{purple}{t} + \textcolor{purple}{t}^2\\$ $\textcolor{purple}{t}^2 + 10\textcolor{purple}{t} - 0.1 = 0\\$ $\textcolor{purple}{t} = \dfrac{-10 ± \sqrt{100 + 0.4}}{2} = 0.00999...\text{ hrs} = \textcolor{purple}{35.96}\text{ seconds} ≈ \textcolor{purple}{36}\text{ seconds}$

A Level

## 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.

## 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]

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

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

A Level

## 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]

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

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

A Level

## 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 – $15.09\text{ s}$

A Level

A Level

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