# Kinematic Equations

## 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.

**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:

**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**:

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

\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}

**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}

**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

\rArr s=1600\div 3=533 \text{ m} (to the nearest metre)