# everything maths & science

You are here: Home Equations of motion

## Equations of motion

In this section we will look at the third way to describe motion. We have looked at describing motion in terms of words and graphs. In this section we examine equations that can be used to describe motion.

This section is about solving problems relating to uniformly accelerated motion. In other words, motion at constant acceleration.

The following are the variables that will be used in this section:

$v → i =initialvelocity(m·s -1 )att=0sv → f =finalvelocity(m·s -1 )attimetΔx →=displacement(m)t=time(s)Δt=timeinterval(s)a →=acceleration(m·s -2 )$(1)

An alternate convention for some of the variables exists that you will likely encounter so here is a list for reference purposes:

$u →=initialvelocity(m·s -1 )att=0sv →=finalvelocity(m·s -1 )attimets →=displacement(m)$(2)

## Interesting Fact:

Galileo Galilei of Pisa, Italy, was the first to determined the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that objects retain their velocity unless a force – often friction – acts upon them, refuting the accepted Aristotelian hypothesis that objects “naturally” slow down and stop unless a force acts upon them. This principle was incorporated into Newton's laws of motion (1st law).

In this book we will use the first convention.

$v → f =v → i +a →tΔx →=(v → i +v → f ) 2tΔx →=v → i t+1 2a →t 2 v f 2 =v i 2 +2a →Δx →$(3)

The questions can vary a lot, but the following method for answering them will always work. Use this when attempting a question that involves motion with constant acceleration. You need any three known quantities ($v → i$, $v → f$, $Δx →$, t or $a →$) to be able to calculate the fourth one.

Problem solving strategy:

1. Read the question carefully to identify the quantities that are given. Write them down.

2. Identify the equation to use. Write it down!!!

3. Ensure that all the values are in the correct units and fill them in your equation.

## Example 1: Equations of motion

### Question

A racing car is travelling North. It accelerates uniformly covering a distance of 725 m in 10 s. If it has an initial velocity of 10 m·s−1, find its acceleration.

#### Identify what information is given and what is asked for

We are given:

$v → i =10m·s -1 Δx →=725mt=10sa →=?$(4)

#### Find an equation of motion relating the given information to the acceleration

If you struggle to find the correct equation, find the quantity that is not given and then look for an equation that has this quantity in it.

We can use (Reference)

$Δx →=v → i t+1 2a →t 2$(5)

$Δx →=v → i t+1 2a →t 2 725m=(10m·s -1 ×10s)+1 2a →×(10s) 2 725m-100m=(50s 2 )a →a →=12,5m·s -2$(6)

The racing car is accelerating at 12,5 m·s−2North.

## Example 2: Equations of motion I

### Question

A motorcycle, travelling East, starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate

• its acceleration

• its final velocity

• at what time the motorcycle had covered half the total distance

• what distance the motorcycle had covered in half the total time.

#### Identify what information is given and what is asked for

We are given:

$v → i =0m·s -1 (becausetheobjectstartsfromrest.)Δx →=64mt=4sa →=?v → f =?t=?athalfthedistanceΔx →=32m.Δx →=?athalfthetimet=2s.$(7)

All quantities are in SI units.

#### Acceleration: Find a suitable equation to calculate the acceleration

We can use (Reference)

$Δx →=v → i t+1 2a →t 2$(8)

#### Substitute the values and calculate the acceleration

$Δx →=v → i t+1 2a →t 2 64m=(0m·s -1 ×4s)+1 2a →×(4s) 2 64m=(8s 2 )a →a →=8m·s -2 East$(9)

#### Final velocity: Find a suitable equation to calculate the final velocity

We can use (Reference) - remember we now also know the acceleration of the object.

$v → f =v → i +a →t$(10)

#### Substitute the values and calculate the final velocity

$v → f =v → i +atv → f =0m·s -1 +(8m·s -2 )(4s)=32m·s -1 East$(11)

#### Time at half the distance: Find an equation to calculate the time

We can use (Reference):

$Δx →=v → i +1 2a →t 2 32m=(0m·s -1 )t+1 2(8m·s -2 )(t) 2 32m=0+(4m·s -2 )t 2 8s 2 =t 2 t=2,83s$(12)

#### Distance at half the time: Find an equation to relate the distance and time

Half the time is 2 s, thus we have $v → i$, $a →$ and $t$ - all in the correct units. We can use (Reference) to get the distance:

$Δx →=v → i t+1 2at 2 =(0m·s -1 )(2s)+1 2(8m·s -2 )(2s) 2 =16mEast$(13)

## Exercise 1: Equations of motion

A car starts off at 10 m·s−1 and accelerates at 1 m·s−2 for 10 s. What is its final velocity?

We use the equations of motion. We are given the initial velocity, acceleration and the time taken. We are asked to find the final velocity.

 ${v}_{f}={v}_{i}+at$ ${v}_{f}=10+\left(1\right)\left(10\right)$ ${v}_{f}=20m\cdot {s}^{-1}$

A train starts from rest, and accelerates at 1 m·s−2 for 10 s. How far does it move?

We are given the initial velocity ($0m\cdot {s}^{-1}$ , since it starts from rest). We also are given the time and the acceleration. We want the distance.

 $\Delta x={v}_{i}\cdot t+\frac{1}{2}a{t}^{2}$ $\Delta x=0+\frac{1}{2}1\left({10}^{2}\right)$ $\Delta x=50m$

A bus is going 30 m·s−1 and stops in 5 s. What is its stopping distance for this speed?

The final velocity is $0m\cdot {s}^{-1}$ , since the bus stops. The initial velocity and time are given.

The acceleration is given by:

 ${v}_{f}={v}_{i}+at$ $a=\left(\frac{{v}_{f}-{v}_{i}}{t}\right)$ $a=\left(\frac{0-30}{5}\right)$ $a=-6m\cdot {s}^{-2}$ $\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ $\Delta x=\left(30\right)\left(5\right)+\left(\frac{1}{2}\right)\left(-6\right){\left(5\right)}^{2}$ $\Delta x=75m$

A racing car going at 20 m·s−1 stops in a distance of 20 m. What is its acceleration?

The final velocity is $0m\cdot {s}^{-1}$  since the car stops. The final velocity and distance are given.

 $vf2=vi2+2a\Delta x$ $0={\left(20\right)}^{2}+2a\left(20\right)$ $0=400+40a$ $-40a=400$ $a=-10m\cdot {s}^{-2}$

The acceleration is negative which indicates that the racing car is slowing down.

A ball has a uniform acceleration of 4 m·s−2. Assume the ball starts from rest. Determine the velocity and displacement at the end of 10 s.

We are given the acceleration and time. The initial velocity is $0m\cdot {s}^{-1}$.

The final velocity is:

 ${v}_{f}={v}_{i}+at$ ${v}_{f}=0+\left(4\right)\left(10\right)$ ${v}_{f}=40m\cdot {s}^{-1}$

The displacement is:

 $\Delta x=\left(\frac{{v}_{f}+{v}_{i}}{2}\right)t$ $\Delta x=\left(\frac{0+40}{2}\right)\left(10\right)$ $\Delta x=200m$

A motorcycle has a uniform acceleration of 4 m·s−2. Assume the motorcycle has an initial velocity of 20 m·s−1. Determine the velocity and displacement at the end of 12 s.

We are given the acceleration, the initial velocity and the time.

The final velocity is:

 ${v}_{f}={v}_{i}+at$ ${v}_{f}=20+\left(4\right)\left(12\right)$ ${v}_{f}=68m\cdot {s}^{-1}$

The displacement is:

 $\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ $\Delta x=\left(20\right)\left(12\right)+\left(\frac{1}{2}\right)\left(4\right){\left(12\right)}^{2}$ $\Delta x=528m$

An aeroplane accelerates uniformly such that it goes from rest to 144 m·hr−1 in 8 s. Calculate the acceleration required and the total distance that it has travelled in this time.

We are given the time, the initial velocity ($0m\cdot {s}^{-1}$ , since it starts from rest) and the final velocity. We must convert the final velocity into $m\cdot {s}^{-1}$.

$\frac{144km}{1hour}=\frac{144×1000}{60×60×1}=40m\cdot {s}^{-1}$

The acceleration is:

 ${v}_{f}={v}_{i}+at$ $40=0+8a$ $a=5m\cdot {s}^{-2}$

The distance travelled in this time is:

 $\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$ $\Delta x=\left(0\right)\left(8\right)+\frac{1}{2}\left(5\right){\left(8\right)}^{2}$ $\Delta x=160m$

## Extension: Finding the equations of motion

The following does not form part of the syllabus and can be considered additional information.

### Derivation of (Reference)

According to the definition of acceleration:

$a →=Δv → t$(14)

where $Δv →$ is the change in velocity, i.e. $Δv=v → f$ - $v → i$. Thus we have

$a →=v → f -v → i tv → f =v → i +a →t$(15)

### Derivation of (Reference)

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For uniformly accelerated motion the most complicated velocity vs. time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of $v → i$, accelerating to a final velocity $v → f$ over a total time t.

To calculate the final displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.

$Area △ =1 2b×h=1 2t×v f -v i =1 2v f t-1 2v i t$(16)
$Area □ =ℓ×b=t×v i =v i t$(17)
$Displacement=Area □ +Area △ Δx →=v i t+1 2v f t-1 2v i tΔx →=v i +v f 2t$(18)

### Derivation of (Reference)

This equation is simply derived by eliminating the final velocity $v f$ in (Reference). Remembering from (Reference) that

$v → f =v → i +a →t$(19)

then (Reference) becomes

$Δx →=v → i +v → i +a →t 2t=2v → i t+a →t 2 2Δx →=v → i t+1 2a →t 2$(20)

### Derivation of (Reference)

This equation is just derived by eliminating the time variable in the above equation. From (Reference) we know

$t=v → f -v → i a$(21)

Substituting this into (Reference) gives

$Δx →=v → i v → f -v → i a+1 2av → f -v → i a → 2 =v → i v → f a →-v → i 2 a →+1 2a →v → f 2 -2v → i v → f +v → i 2 a → 2 =v → i v → f a →-v → i 2 a →+v → f 2 2a →-v → i v → f a →+v → i 2 2a →2a →Δx →=-2v → i 2 +v → f 2 +v → i 2 v → f 2 =v → i 2 +2a →Δx →$(22)

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

## Applications in the real-world

What we have learnt in this chapter can be directly applied to road safety. We can analyse the relationship between speed and stopping distance. The following worked example illustrates this application.

### Example 3: Stopping distance

#### Question

A truck is travelling at a constant velocity of 10 m·s−1 when the driver sees a child 50 m in front of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of −1,25 m·s−2. His reaction time to hit the brakes is 0,5 seconds. Will the truck hit the child?

##### Analyse the problem and identify what information is given

It is useful to draw a time-line like this one:

We need to know the following:

• What distance the driver covers before hitting the brakes.

• How long it takes the truck to stop after hitting the brakes.

• What total distance the truck covers to stop.

##### Calculate the distance $AB$

Before the driver hits the brakes, the truck is travelling at constant velocity. There is no acceleration and therefore the equations of motion are not used. To find the distance travelled, we use:

$v=d t10=d 0,5d=5m$(23)

The truck covers 5 m before the driver hits the brakes.

##### Calculate the time $BC$

We have the following for the motion between $B$ and $C$:

$v → i =10m·s -1 v → f =0m·s -1 a=-1,25m·s -2 t=?$(24)

We can use (Reference)

$v → f =v → i +at0=10m·s -1 +(-1,25m·s -2 )t-10m·s -1 =(-1,25m·s -2 )tt=8s$(25)
##### Calculate the distance $BC$

For the distance we can use (Reference) or (Reference). We will use (Reference):

$Δx →=(v → i +v → f ) 2tΔx →=10+0 s(8)Δx →=40m$(26)
The total distance that the truck covers is $d AB$ + $d BC$ = 5 + 40 = 45 meters. The child is 50 meters ahead. The truck will not hit the child.