# SUVAT Equations

## SUVAT Equations Revision

**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

**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}

**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]**

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}

**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}

## SUVAT Equations Example Questions

**Question 1:** Luke drops a ball from a height of 15\text{ m}. What velocity does the ball hit the ground at?

**[1 mark]**

Take g = 9.8\text{ ms}^{-2}.

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}

**Question 2:** A plane is coming in to taxi at an airport. It lands parallel to the surface at 480\text{ ms}^{-1}, and takes exactly one minute to come to a complete stop. What is its acceleration?

**[2 marks]**

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)

**Question 3: **James, Jeremy and Richard are having a quarter-mile race. From rest, Richard accelerates at a constant rate of 8\text{ mph/s}. Jeremy sets off one second earlier than he should, and accelerates at 7\text{ mph/s}. James is given a 3 \text{ second} head start on the other two, but his acceleration is only 5.5\text{ mph/s}. Who crosses the line first? Who’s last?

**[3 marks]**

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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