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Chapter 8. equations of motion (part 2)

# Chapter 8. equations of motion (part 2)

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back to dingo

Previously ...

Dear Dingo,

Really? You didn’t think

I’d see the crane?!

Sincerely not yours,

Dear Dingo,

You didn’t think I’d see

the crane?!

Sincerely not yours,

Emu.

Emu.

Now ACME has an amazing new cage launcher

But you can’t keep the Dingo down for long - especially when ACME has an

amazing new cage launcher! Once installed, it’ll propel a standard ACME

cage straight up in the air at a speed of your choice.

It’s ideal for a more subtle approach - you can launch the cage from ground

level, instead of having a big crane that the Emu will spot.

You just need to work out what velocity to launch it at so that it lands back on

the target when the Emu arrives, exactly 2.0 s after you launched it.

ACME

Cage Launcher

1

2

- Launches a standard

ACME cage straight up

in the air.

- Waterproof

- Variable launch speeds.

- Payment plans and

financing available

284   Chapter 8

I gotta have that cage

launcher! And I’m going

back to my old spot, where

the Emu takes 2.0 s to arrive.

equations of motion (part 2)

So we have to launch the cage

up in the air so that it lands

2.0 s later. That sounds a lot

harder than just dropping it.

Joe: Hmm, could we try using the equation we worked out in

chapter 7? Y’know, x = x0 + v0t + ½at2.

Frank: But that was for a falling thing. This time, the cage is

going up. It’s not the same thing.

Joe: But once the cage gets to its maximum height, it falls back down

again. So it’s kinda the same. Or, well, at least the falling down part

of it is!

Frank: But what about the first half, when the cage is going up?

Jim: Actually, the equation might work OK then too. It’s supposed

to work in any situation where the acceleration’s constant, right?

And I think that the acceleration due to gravity is constant, whatever

direction you’re moving in.

Frank: But how can the cage be accelerating downwards, when it’s

moving upwards?

Joe: Acceleration is rate of change of velocity, right? And the

cage gets slower as it goes up. So the acceleration vector must be

pointing downwards, or else it wouldn’t get slower.

Always start

with a sketch!

Frank: I think I’d find it easier to visualize with a sketch ...

Draw a sketch of the cage just after it’s been

launched straight up in the air. Mark on the initial

velocity vector v0 , the acceleration vector a, and any

other information you know about the problem.

Do you think it’s going to be OK to reuse the

equation x = x0 + v0t + ½at2 in this new scenario?

Why / why not?

you are here 4   285

is acceleration constant?

Draw a sketch of the cage just after it’s been

launched straight up in the air. Mark on the initial

velocity vector v0, the acceleration vector a, and any

other information you know about the problem.

Cage takes t = 2.0 s

to go up then down.

The cage starts

and ends at the

same point, so x0

and x are both 0 m.

Do you think it’s going to be OK to reuse the

equation x = x0 + v0t + ½at2 in this new scenario?

Why / why not?

v0 = ?

x0 = 0 m

x=0m

a = 9.8 m/s2

The acceleration vector points down regardless

of the direction of the cage’s velocity.

The acceleration will be constant at 9.8 m/s2

downwards. The equation’s supposed to be used

for constant acceleration, so it’ll probably work.

Give your hand-drawn acceleration vectors doublearrowheads to distinguish them from velocity vectors.

The acceleration due to gravity is constant

The equation of motion you worked out last time,

x = x0 + v0t + ½at2, is supposed to be OK in any

situation where the acceleration is constant.

v = 0 m/s

If you can reuse this equation, it’ll be much much

faster than going through the rigmarole of trying to

design and carry out an experiment where you shoot

things straight up into the air.

If an object is

v

acted on ONLY BY

GRAVITY, it has

an acceleration

of 9.8 m/s2

a = 9.8 m/s

downwards,

whatever its

When the cage is going up, it

velocity is.

gets slower. This is because it’s

2

a = 9.8 m/s2

When the cage’s

velocity is zero, its

acceleration due to

gravity is still 9.8 m/s2.

When the cage is going

down, it gets faster.

This is because it’s being

accelerated downwards

at a rate of 9.8 m/s2

Although you originally

worked out this equation

from a graph you plotted by

dropping things down from

a height, any object that’s

accelerated only by gravity has

a constant acceleration of

9.8 m/s2 downwards. It

doesn’t matter whether its

velocity vector points up, down,

sideways, or at an angle.

being accelerated downwards

at a rate of 9.8 m/s2

286   Chapter 8

v

a = 9.8 m/s2

equations of motion (part 2)

reality, it should work with

these numbers in it.

t = 2.0 s

v0 = ?

x0 = 0 m

x=0m

a = 9.8 m/s2

I’m not convinced! If the cage starts

and finishes in the same place, then x

and x0 are both zero. So how does the rest of

the equation work? How can two terms added

together be zero?!

Good thinking - try to imagine your

equation with some numbers in it.

The cage starts and finishes at ground level.

This means that both x and x0 are zero, and

your equation becomes 0  = 0 + v0t + ½at2

when you put these values in.

So for the equation to be true, the two terms on

the right hand side, v0t and ½at2, must add up

to zero. But how can two terms added together

be zero? It’s time to ...

BE the equation

Your job is to imagine you’re the equation. Here, both x and x0 are 0 because the

cage starts and finishes in the same place. This means that the left hand side of

the equation = 0, and on the right hand side, there are two non-zero terms

ADDED together. Is there any way this is possible, or will we need to

work out a different equation for the cage?

x = x0 + v0t + 21 at2

you are here 4   287

vectors have direction

BE the equation - SOLUTION

Your job is to imagine you’re the equation. Here, both x and x0 are 0 because the

cage starts and finishes in the same place. This means that the left hand side of

the equation = 0, and on the right hand side, there are two non-zero terms

ADDED together? Is there any way this is possible, or will we need to

work out a different equation for the cage?

x = x0 + v0t + 21 at2

Velocity and acceleration are vectors.

v0 is upwards, and a is downwards - as they point in opposite

directions, they’ll have opposite signs.

It IS possible for a positive number and a negative number to equal

zero when you add them together. So the equation could be OK.

v0 is UPWARDS.

v0 = ?

a = 9.8 m/s2

a is DOWNWARDS.

Velocity and acceleration are in opposite

directions, so they have opposite signs

When x and x0 are both zero, your equation becomes

0 = 0 + v0t + ½at2. So the terms on the right hand side

You’re interested in what’s happening at t = 2.0 s, so both

t and t2 are positive. Therefore, the signs of the terms

v0t and ½at2 are determined by the signs of v0 and a.

v0

Vectors have

DIRECTION!

You’re working with vectors! So as well as having a

size, v0 and a have a direction. The acceleration due

to gravity, a, always acts downwards. But the initial

velocity, v0, is upwards. So v0 and a have opposite

signs. One is positive and the other is negative.

As v0 and a have opposite signs, it’s perfectly reasonable

to say that there’s a certain value for v0 where

v0t + ½at2 = 0. And that’s the value you want to work

out, as it’s the launch velocity for the cage!

a

As v0 and a point in

opposite directions, they

will have opposite signs.

288   Chapter 8

You use positive

and negative

signs to show

the direction.

equations of motion (part 2)

+

Do we get to decide which direction is

positive and which is negative? So this is just

another way to use vectors?

Yes. With vectors that point in opposite

directions, you get to decide which

way is positive and which is negative.

Vectors are vectors, so as long as you are consistent

and dealing with opposites, you get to choose

whether to make up or down the positive direction.

When something’s moving through the air, it’s usually

easiest to make up the positive direction and

down the negative direction.

-

It’s conventional to make up

the positive direction and

down the negative direction.

But when we were dropping the cage, we said

that down was positive. Why suddenly change

now to make up positive - it’s confusing!

Choose the direction that makes

the math easier.

It’s easy to ‘lose’ or forget about minus signs

when you’re doing calculations and end up with

the wrong answer. When you dropped the cage,

its displacement, velocity and acceleration all

pointed in the same direction - downwards.

So by making down the positive direction, you

didn’t have to deal with minus signs.

But now that the cage is going up then down,

with v0 and a pointing in opposite directions,

you’ve nothing to gain by making down positive.

You’re better off sticking with the convention

of making up positive so that when you draw a

displacement-time graph, up on the graph is the

same direction as up in the real world.

You’re less prone

to making mistakes

when doing math

with only positive

numbers than you

are when there

are negative

numbers around.

you are here 4   289

opposite directions opposite signs

Q:

I still don’t get how you can add two numbers together

A:

Numbers can be negative as well as positive. If one of the

numbers is negative, it can work out like that. For example, if

v0t = -2 and ½at2 = 2, then you have -2 + 2 = 0.

Q:

Sorry ... I still don’t quite see how two numbers added

together can be zero?

A:

Suppose you spend \$10 entering a competition, then win a

\$10 prize in it. You’ve earned (-10) + 10 = 0 dollars. The sum of

the (negative) entry fee and the (positive) prize win comes out to

a zero balance.

Q:

Why would I want to add a negative number to a

positive number when I can just do a subtraction?

A:

Because when you have an equation like x = x0 + v0t + ½at2,

you don’t know in advance which variables are positive and which

are negative. But when you put the numbers in, it’ll work out as

long as you get the minus signs right.

Q:

Q:

What would happen if I made down the positive

A:

The math would all still work out, as long as you make sure

you’re consistent. You just have to be careful to do the right thing

when you’re adding or subtracting a negative number.

Q:

How would I figure out if I’d made a mistake with the

minus signs?

A:

You can see if your answer SUCKs. If a minus sign has

gone astray, then the answer may end up a very different size

from what you’d expect. So make sure you have a rough idea of

compare it with the result of your calculation.

Q:

OK, so I think I have the negative numbers, vectors and

directions all figured out. Is there anything else I should do?

A:

As you’ve never used this equation before to deal with

something going up then down, it wouldn’t hurt to sketch some

graphs to confirm that it’s going to be OK ...

So the variables x0, x, v0, v and a could all be negative

because they’re vectors?

A:

Yes, vectors have both a size and a direction. When all of

your vectors lie along one line (like in this case, they’re either

pointing up or down) then you can choose to make one direction

positive and the other direction negative.

Q:

And I get to choose whatever direction I want to be

positive? Either up or down?

A:

Yes, as long as you’re consistent throughout. But it’s usual

to make up the positive direction so that when you plot a graph,

up on the graph corresponds to up in real life.

Vectors in opposite directions

have opposite signs.

290   Chapter 8

I only get one shot at this.

Are you sure this equation’s

gonna work? Can you draw me a

graph or something?

equations of motion (part 2)

OK, so we’ve got an equation.

But does it look right?!

Frank: I guess we need to do some graphs that show it’s OK to

use the equation to deal with the cage going straight up in the air

and back down again.

Jim: But that’s gonna be difficult. We usually draw graphs of

experimental results, but I don’t think we can do that this time.

We only get one shot at launching the cage, and if we miniaturize

the experiment, we can’t really measure the launch speed.

Joe: Hmmm. Maybe we could sketch graphs of what we know

happens to something that goes up and down. Then put some

numbers into the equation and plot it to see if it comes out the same

shape - like doing that GUT check thing - comparing the equation

with the graph by trying out some values.

Frank: But how do we sketch a displacement-time graph for

something that goes up then down? Is it curved? Is it straight? Does

the shape depend on whether it’s going up or down? I don’t think

we can do that straight off.

Joe: We could start off with the acceleration-time graph. We

know that has a constant value of 9.8 m/s2.

Jim: You mean -9.8 m/s2 ... up is positive, so down is negative!

Equations represent reality.

If you sketch a graph

of what happens in real

life, it should be the same

shape as the graph for

Joe: Yeah, well, acceleration is rate of change of velocity. So the

value of the acceleration is the slope of the velocity-time graph.

The value of the acceleration is constant: -9.8 m/s2, so the slope of

the velocity-time graph must also be constant at -9.8 m/s2.

Frank: Err, a negative slope?! What would that look like?!

Jim: I guess it would go the other way, down from left to right

Joe: That sounds about right. Then when we have the velocitytime graph, we can use the fact that velocity is rate of change

of displacement. So the value of the velocity is the slope of the

displacement-time graph.

Frank: Which lets us draw a displacement-time graph as well.

Jim: Great!

here - the important thing

is the SHAPES of the graphs.

What might the velocity-time and

displacement-time graphs look like?

you are here 4   291

negative and positive slope

The same principle applies to the

slope of a displacement-time graph

and the value of the velocity.

Slope Up Close

Since velocity is a vector, it can be either positive or negative. The slope of a

velocity-time graph shows you the value of the acceleration.

If the change in velocity is positive, then the graph slopes up, and the

acceleration is positive. If the change in velocity is negative, then the graph

slopes down, and the acceleration is negative.

Graph of velocity vs. time

v

Acceleration.

∆v

is positive.

∆v

∆v ∆t

Sloping

UP

∆t

∆t

is positive.

t

is positive.

a = ∆v

∆t

Slope of velocity-time

graph is positive, so value

of acceleration is positive,

i.e., it’s accelerating in this

direction.

If a graph goes up, its slope is positive.

Graph of velocity vs. time

v

∆v

Sloping

DOWN

is negative.

Acceleration.

∆v

∆t

is negative.

a = ∆v

∆t

∆v

∆t

∆t

t

is positive.

Slope of velocity-time

graph is negative, so value

of acceleration is negative,

i.e., it’s accelerating in this

direction.

If a graph goes down, its slope is negative.

292   Chapter 8

thinking about values and slopes. What

you’re doing here isn’t too different.

You can use one graph to work out

the shapes of the others

equations of motion (part 2)

Graphs for Alex the cyclist

(constant velocity)

Acceleration, velocity, and displacement are all related to each

other. If you have the graph of one of them and some initial

values for the others, you can use these to sketch the shapes of

the other graphs.

Graphs for a falling object

(constant acceleration)

Displacement-time

x

x

Displacement-time

Constant

Increasing

t

Velocity is constant

at all times, so always

has the same value.

So the velocity-time

graph is a flat line.

You know that the launched cage has a constant acceleration

of -9.8 m/s2. Since the value of the acceleration is constant

and negative, the slope of the velocity-time graph must be

constant and negative too.

v

Constant

value.

t

a

t

Velocity-time

v

Velocity-time

Increasing

value.

Zero

t

Acceleration-time

a

Acceleration-time

Constant

value.

Zero

value.

Velocity is constant,

so acceleration is zero.

Constant

t

t

Use the value of your acceleration-time graph to sketch the shape of the velocity-time

graph. Try to imagine the velocity of the cage as it goes up then back down again. The

first dotted line is the top of its flight, and the second one is when it lands again.

Don’t

Velocity (m/s)

displacement-time

the

Don’t worry

graph for the moment

you’ll do that next.

happens after

it lands.

Graph of velocity vs. time for the cage

which goes up then down again

v0

We’ve put v0

on to help

you out.

Time (s)

This is the

velocity vector

at t=0.

Acceleration

(m/s2)

Graph of acceleration vs. time for the cage

which goes up then down again

This graph shows that the

acceleration is constant, with this

size and this direction at all times.

These two statements are the same

thing said different ways around.

Time (s)

a

-9.8

Acceleration (a) is

constantly -9.8 m/s2.

Top of

flight

The slope of

the velocity-time

graph gives the value

of the acceleration.

Landing

The value of the

acceleration gives

the slope of the

velocity‑time graph.

you are here 4   293

vectors have size and direction

Use the value of your acceleration-time graph to sketch the shape of the velocity-time

graph. Try to imagine the velocity of the cage as it goes up then back down again. The

first dotted line is the top of its flight, and the second one is when it lands again.

Velocity (m/s)

Graph of velocity vs. time for the cage

which goes up then down again

The value of the velocity

is zero when the cage is at

the top of its flight path.

v0

We’ve put v0

on to help

you out.

When the cage is falling down,

its velocity is negative, as

down is the negative direction.

Time (s)

The slope is negative, so

the graph slopes down.

Acceleration

(m/s2)

Graph of acceleration vs. time for the cage

which goes up then down again

When the cage comes back

down again, it gets faster

and faster - but in the

downwards DIRECTION.

The velocity-time graph has a constant slope of

-9.8 m/s2 because the acceleration-time graph

has a constant value of -9.8 m/s2.

Time (s)

-9.8

v

Acceleration (a) is

constantly -9.8 m/s2.

Velocity-time graph

When x is positive,

vector arrows point up.

When x is negative,

vector arrows point down.

t

Tails of vectors

are at v = 0.

Graph shows you where

Vectors are arrows, right?

So what have the lines on the

graphs got to do with vectors?

The graphs show the size and

direction of a vector at any time

The line on a graph shows the size and

direction of the quantity you’re plotting

at any time. If you imagine the tail of the

arrow on the horizontal axis, the head will be

just touching the line.

294   Chapter 8