Albert Teen
powered by
Albert logo

YOU ARE LEARNING:

pointer
Uniform Acceleration
lessonMenuButton

Uniform Acceleration

lesson introduction

Uniform Acceleration

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

1

What do you think uniform means?

hint button
2

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

hint button
3

What is this cyclist's acceleration?

hint button
block image
4

That means that the cyclist's velocity increases by 0.5 m/s0.5 \space m/s every...

hint button
block image
5

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

For example, this cyclist's speed will accelerate by 0.5 m/s0.5 \space m/s every second.

block image

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

v2u2=2asv^2-u^2=2as

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

v2u2=2asv^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 aa stand for?

hint button
2

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

hint button
3

So far we've defined aa as uniform acceleration, uu as initial velocity and vv as final velocity. What does that make ss?

hint button
4

So we have aa for uniform acceleration, uu for initial velocity, vv for final velocity and ss for distance.

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

Calculating acceleration, aa

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?

hint button
block image
2

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

u=2m/su=2\:m/s

v=5m/sv=5\:m/s

s=6ms=6\:m

a=?a=\:?

block image
3

Now we need to rearrange our equation v2u2=2asv^2-u^2=2as to make aa the subject. What will this look like?

hint button
block image
4

Great, now that we have all the information we need, calculate the acceleration using a=v2u22sa=\frac{v^2-u^2}{2s}.

hint button
block image
1

Use a=v2u22sa=\frac{v^2-u^2}{2s} to work out this car's acceleration. Give your answer to 2 decimal places.

hint button
block image

A runner starts a race and uniformly accelerates at 0.5m/s20.5\:m/s^2 until she reaches a velocity of 3m/s3\:m/s.

1

What information is missing from the above statement?

hint button
2

We know that we are going to be asked to find the distance the runner ran. Write v2u2=2asv^2-u^2=2as where ss is the subject.

hint button
3

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

hint button

Calculating the final velocity, vv

1

What values are given in this picture?

hint button
block image

You can select multiple answers

2

So we are missing vv. How do we rearrange v2u2=2asv^2-u^2=2as to make the final velocity the subject?

hint button
block image
3

Now use v2=u2+2asv^2=u^2+2as to work out v2v^2 for this cyclist.

hint button
block image
4

So v2=81 m/sv^2=81 \space m/s. How do you work out the cyclist's final velocity, vv then?

hint button
block image
5

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

v=u2+2asv=\sqrt{u^2+2as}

block image
1

Use v=u2+2asv=\sqrt{u^2+2as} to work out this train's final velocity.

hint button
block image
1

What is this person's final velocity?

hint button
block image

A bowling ball is dropped from a height of 5m5m and accelerates uniformly due to gravity at 10m/s210\:m/s^2. It reaches a velocity of 10m/s10\:m/s right before it hits the ground. What was its initial velocity?

hint button

Summary!

1

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

The equation v2u2=2asv^2-u^2=2as can only be used for uniformly accelerated motion.

block image
2

To calculate acceleration, aa, we divide both sides by 2s2s to get...

a=v2u22sa=\frac{v^2-u^2}{2s}

block image
3

To calculate distance, ss, we divide both sides by 2a2a to get...

$s=\frac{v^2-u^2}{2a}$ 
block image
4

To calculate the final velocity, vv we +u2+u^2 to both sides and take the square-root to get...

v=u2+2asv=\sqrt{u^2 + 2as}

block image
5

To calculate the initial velocity, uu, we rearrange to get...

u=v22asu=\sqrt{v^2-2as}

block image