Albert Teen

YOU ARE LEARNING:

Uniform Acceleration

# Uniform Acceleration

### Uniform Acceleration

Distance and acceleration can be related to each other in a single equation.

1

What do you think uniform means?

2

So if an object has uniform acceleration, what does that mean?

3

What is this cyclist's acceleration?

4

That means that the cyclist's velocity increases by $0.5 \space m/s$ every...

5

"Uniform acceleration" is the same as "constant acceleration"

For example, this cyclist's speed will accelerate by $0.5 \space m/s$ every second.

For uniform acceleration, we can use the following equation that links velocity, distance and acceleration.

$v^2-u^2=2as$

This lesson will teach you how and when to use this.

$v^2-u^2=2as$

1

First of all, with any equation, we have to define the variables. Starting with the most obvious one, what does $a$ stand for?

2

The equation for acceleration that we've already looked at is $a=\frac{\Delta v}{t}$. This can also be written like this $a=\frac{v-u}{t}$. What do you think $v$ and $u$ stand for, respectively?

3

So far we've defined $a$ as uniform acceleration, $u$ as initial velocity and $v$ as final velocity. What does that make $s$?

4

So we have $a$ for uniform acceleration, $u$ for initial velocity, $v$ for final velocity and $s$ for distance.

Let's now look at how to use this equation.

Calculating acceleration, $a$

1

With any question, the first thing you have to do is write down all the variables the question gives you. In this case, what variables do you know?

2

Ok, so now we have all of our variables defined.

$u=2\:m/s$

$v=5\:m/s$

$s=6\:m$

$a=\:?$

3

Now we need to rearrange our equation $v^2-u^2=2as$ to make $a$ the subject. What will this look like?

4

Great, now that we have all the information we need, calculate the acceleration using $a=\frac{v^2-u^2}{2s}$.

1

Use $a=\frac{v^2-u^2}{2s}$ to work out this car's acceleration. Give your answer to 2 decimal places.

A runner starts a race and uniformly accelerates at $0.5\:m/s^2$ until she reaches a velocity of $3\:m/s$.

1

What information is missing from the above statement?

2

We know that we are going to be asked to find the distance the runner ran. Write $v^2-u^2=2as$ where $s$ is the subject.

3

Calculate how far the runner travelled until she reached her final velocity.

Calculating the final velocity, $v$

1

What values are given in this picture?

2

So we are missing $v$. How do we rearrange $v^2-u^2=2as$ to make the final velocity the subject?

3

Now use $v^2=u^2+2as$ to work out $v^2$ for this cyclist.

4

So $v^2=81 \space m/s$. How do you work out the cyclist's final velocity, $v$ then?

5

So you worked out this cyclists final velocity using ...

$v=\sqrt{u^2+2as}$

1

Use $v=\sqrt{u^2+2as}$ to work out this train's final velocity.

1

What is this person's final velocity?

A bowling ball is dropped from a height of $5m$ and accelerates uniformly due to gravity at $10\:m/s^2$. It reaches a velocity of $10\:m/s$ right before it hits the ground. What was its initial velocity?

Summary!

1

An object that has uniform acceleration has constant change in velocity.

The equation $v^2-u^2=2as$ can only be used for uniformly accelerated motion.

2

To calculate acceleration, $a$, we divide both sides by $2s$ to get...

$a=\frac{v^2-u^2}{2s}$

3

To calculate distance, $s$, we divide both sides by $2a$ to get...

$s=\frac{v^2-u^2}{2a}$

4

To calculate the final velocity, $v$ we $+u^2$ to both sides and take the square-root to get...

$v=\sqrt{u^2 + 2as}$

5

To calculate the initial velocity, $u$, we rearrange to get...

$u=\sqrt{v^2-2as}$