Kinematic Equations

GCSELevel 4-5OCR

Kinematic Equations Revision

Kinematic Equations

Most of the formulas that are used at GCSE are for calculating measures in geometry, for example area and volume. However, you will need to know some of the kinematic formulas that are used to solve problems that involve moving objects.

The variables used in the formulas are:

Displacement $(\bold{s})$

Initial Velocity $(\bold{u})$

Final Velocity $(\bold{v})$

Acceleration $(\bold{a})$

Time $(\bold{t})$

Make sure you are happy with the following topics before continuing.

Rearranging Formulas

Key Equations

There are $3$ formulas you will need to be able to work with and rearrange when required:

$\textcolor{#bd0000}{v}=\textcolor{#f95d27}{u}+\textcolor{#2730e9}{a}\textcolor{#00bfa8}{t}$

$\textcolor{#aa57ff}{s}=\textcolor{#f95d27}{u}\textcolor{#00bfa8}{t}+\dfrac{1}{2}\textcolor{#2730e9}{a}\textcolor{#00bfa8}{t}^2$

$\textcolor{#bd0000}{v}^2=\textcolor{#f95d27}{u}^{2}+2\textcolor{#2730e9}{a}\textcolor{#aa57ff}{s}$

You will usually be given $3$ or $4$ values and be expected to substitute these in to find a desired value.

Note:

The standard units of measure for the variables are:

Metres $(\textcolor{red}{\text{m}})$ for displacement

Metres per second $(\textcolor{blue}{\text{m/s}} \text{ or } \textcolor{blue}{\text{ms}^{-1}})$ for velocity

Metres per second per second $(\textcolor{green}{\text{m/s}^2} \text{ or } \textcolor{green}{\text{ms}^{-2}})$ for acceleration

Seconds $(\textcolor{purple}{\text{s}})$ for time

Example 1: Substituting Values

a) Using the formula $s=ut+\dfrac{1}{2}at^2$ , find $s$ when $u=5 \text{ m/s}$ , $t=6 \text{ s}$ and $a=9.8 \text{ m/s}^2$

[2 marks]

b) Using the formula $v^2=u^2+2as$ , find $v$ when $u=0 \text{ m/s}$ , $a=3.8 \text{ m/s}^2$ and $s=140 \text{ m}$

Give your answer to $1$ decimal place.

[2 marks]

a) Substituting in the values given:

\begin{aligned} s &=5\times6+\dfrac{1}{2}\times 9.8 \times 6^2 \\ &=206.4 \text{ m}\end{aligned}

b) Substituting in the values given:

$v^2 =0^2+2\times 3.8 \times 140$

$\rArr v^2=1064 \text{ m/s}^2$

$\rArr v=\sqrt{1064}=32.6 \text{ m/s}^2$ to $1$ decimal place.

Example 2: Rearranging Equations

a) Given that $v=22 \text{ m/s}$ , $u=4\text{ m/s}$ and $t=12\text{ s}$ , find the value of $a$ using the formula $v=u+at$

[3 marks]

b) Given that $s=90\text{ m}$ , $t=16\text{ s}$ and $a=-9.8\text{ m/s}^2$ , find the value of $u$ using the formula $s=ut+\dfrac{1}{2}at^2$

[3 marks]

a) There are two methods to find $a$ , we could either rearrange the equation first and then substitute the given values in to find $a$ or substitute first and then rearrange. For this question we will rearrange then substitute:

$v=u+at$

$\rArr v-u=at$

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

$a=\dfrac{22-4}{12}=1.5 \text{ m/s}^2$

b) As this is quite a complicated equation, we will substitute and then rearrange:

90=16u+\dfrac{1}{2}\times (-9.8)\times 16^2

Simplifying and rearranging we get:

16u=1344.4 \text{ m/s}

\rArr u=\dfrac{1344.4}{16}=84.025\text{ m/s}

Example 3: Interpreting Real World Scenarios

For the following question you may use the formula $v^2=u^2+2as$

A car is travelling at $32 \text{ m/s}$ . The driver of the car then brakes, as they see a traffic light has turned red $120 \text{ m}$ away. The car decelerates at $6 \text{ m/s}^2$ .

Does the car stop before it reaches the traffic lights?

Show your working.

[4 marks]

To answer this question we need to find out how far the car will travel before it stops, or in other words, before $v=0 \text{ m/s}$

So we will need to substitute the following values into $v^2=u^2+2as$

$u=32 \text{ m/s}$

$v=0 \text{ m/s}$

$a=-6 \text{ m/s}^2$

Then rearrange to find $s$

0^2=32^2+2\times (-6)\times s

\rArr 12s=1024

$\rArr s=\dfrac{1024}{12}=85.3 \text{ m}$ (to $1$ decimal place)

Therefore the car will stop in time, as the traffic lights are $90 \text{ m}$ away.

Note: Make sure to use $a=-6 \text{ m/s}^2$ instead of $a=6 \text{ m/s}^2$ as the car is decelerating in this scenario.

Kinematic Equations Example Questions

Question 1: You are given that $s=65 \text{ m}$ , $u=14 \text{ m/s}$ and $a=3 \text{ m/s}^2$

Using $v^2=u^2+2as$ , find $v$ .

Give your answer to $1$ decimal place.

[2 marks]

Substituting in the given values:

$v^2=14^2+2\times 3 \times 65$

$\rArr v^2=586 \text{ m/s}$

$\rArr v=\sqrt{586}=24.2 \text{ m/s}$

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Question 2: You are given that $v=24 \text{ m/s}$ , $u=6 \text{ m/s}$ and $a=3.4 \text{ m/s}^2$

Using $v=u+at$ , find the values of $t$ .

Give your answer to $1$ decimal place.

[2 marks]

Substituting the values given:

$24=6+3.4t$

$\rArr 3.4t=18$

$\rArr t=\dfrac{18}{3.4}=5.3 \text{ s}$

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Question 3: A train accelerates from stand still at a station with a uniform acceleration of $1.5 \text{ m/s}^2$ until it reaches a velocity of $40 \text{ m/s}$ .

Calculate how far the train travels while it accelerates to a velocity of $40 \text{ m/s}$ .

Use the equation $v^2=u^2+2as$

Give your answer to the nearest metre.

[3 marks]

To solve real world problems like this, it is often useful to write down what variables we are given from the question:

$u=0 \text{ m/s}$

$a=1.5 \text{ m/s}^2$

$v=40 \text{ m/s}$

We need to substitute and rearrange to find $s$ .

Substituting our values in:

$40^2=0^2+2\times 1.5\times s$

$\rArr 1600=3s$