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Chapter 6. Displacement, Velocity, and Acceleration

Chapter 6. Displacement, Velocity, and Acceleration

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what’s going on here?



Just another day in the desert ...

The Dingo pushes the

cage off the platform

as soon as the Emu

rounds the corner.



The Dingo wants the Emu to

stay still for long enough to

deliver an invitation to his

birthday party.



The Dingo needs to

know how high the

platform should be,

and whether the

cage can cope with

falling that far.



The Emu runs at 54

kilometers per hour.



The target is 30 m

from the corner.

The cage will fall

on the target.

204   Chapter 6

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displacement, velocity, and acceleration



... and another Dingo-Emu moment!

Every year it’s the same. The Dingo wants to invite the Emu to his birthday

party - but the daft bird won’t stop running for long enough for him to

deliver the invitation. So this year, the Dingo’s decided that extending a

paw of friendship needs drastic measures. He’s hired a crane, and wants to

push a cage off the platform the moment the Emu rounds the bend. But is

this practical? What height does the platform need to be, and will

the cage be able to handle hitting the ground at a high speed?



Emu - Runningus fasticus



54 kilometers per hour



So the Dingo calls the crane company’s customer service department to ask

some questions ...



Crane Company Magnets



The crane company gets to work on the problem. But we accidentally dropped their memo

and some of the words fell off. Your job is to put them back in the right places. You might

use some magnets more than once, and some not at all.

Also, underline the most important parts in the memo to separate the important stuff

from the fluff - the wheat from the chaff.



To: Dingo



Re: Cage



, tricky! The

g

ding exactly on runnin

lan

e

cag

g

lin

fal

m,

Hmm

- and we

re on the computer up the

is

Emu’s

past the corner. The

get

tar

and

ne

cra

the

up

set

rounds

as the

falls at the same time

the cage falls in the

out the

k

wor

we

If

.

ner

cor

the

to that

,we can set the crane

run

to

Emu

it takes the

. Be careful - the

and take home a fat

less than 25 m/s.

it hits the ground at

is only guaranteed if

30 m

time



Emu



height

54 km/h



cage



commission

distance



speed



54 m/s



velocity

30 km



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magnets solution



Crane Company Magnets - Solution



The crane company gets to work on the problem. But we accidentally dropped their memo

and some of the words fell off. Your job is to put them back in the right places. You might

use some magnets more than once, and some not at all.

Also, underline the most important parts in the memo to separate the important stuff

from the fluff - the wheat from the chaff.



To: Dingo



Re: Cage



, tricky! The

Emu

g

ding exactly on runnin

lan

e

cag

g

lin

fal

m,

Hmm

- and we

54 km/h

re on the computer up the

is

ed

spe

Emu’s

past the corner. The

30 m

get

tar

and

ne

cra

the

up

set

rounds

Emu

as the

falls at the same time

cage

time

the cage falls in the

tance

dis

the

out

the corner. If we work

to that

,we can set the crane

30 m

run

to

Emu

it takes the

. Be careful - the

commission

and take home a fat

height

.

und at less than 25 m/s

teed if it hits the gro

ran

gua

y

onl

is

cage



NOTES

30 km

54 m/s



These didn’t get used

because the units are wrong.



What time does the cage fall for?

What height should the crane be?

Will the cage be going faster than

25  m/s when it hits the ground?



velocity



The Emu’s speed, rather than

his velocity, is important , as

the road is curved.



Which of these would

you try to work out first?



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displacement, velocity, and acceleration



How can you use what you know?

The Dingo drops the cage as soon as the Emu rounds

the corner. Then, the cage falling and the Emu

running both take the same time to reach the target.

The time that the Emu takes to arrive depends on

the speed he runs at and the distance he covers

from the corner to the target. As the Emu always

runs with a constant speed, you already know an

equation you can use to do this.

Once you know the time it takes the Emu to arrive,

you’ll have to figure out how far the cage falls

during that time. This will give the Dingo the height

that he needs to set the platform at.



The cage takes

time to get from

the platform to

the target.



The Emu takes

time to get

from the corner

to the target.



However, if the cage travels faster than 25 m/s in

the time it takes for the Emu to reach the target, this

plan won’t work because the cage will hit the ground

and be destroyed upon impact.



You haven’t dealt with falling

things yet - but don’t worry,

that’s what this chapter’s about!



Hint: You’ll need

to convert units.



These times are equal.



First things first. Work out the time it takes the Emu to cover 30 m from the

corner to the target at a speed of 54 km/h.



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sharpen solution



Work out the time it takes the Emu to cover 30 m from the corner to the target

at a speed of 54 km/h.



Convert units: km/h to m/s

This symbol

means ‘implies 54 km/h in m/s = 54 km x 1000 m

hours

1 km

that’. You

= 15 m/s

can use it

going from

one line to

Work out the time it takes:

the next as

distance

speed =

you rearrange

time

an equation.

speed time =

distance

time =



Rearrange

equation to get

time = ...



distance

speed



=



1 hour x 1 min

x 60

mins

60 s



After stringing together

conversion factors, you’re left

with meters on the top and

seconds on the bottom - m/s.



Equation comes from the units

of speed. Meters per second is

a distance divided by a time.



30 m

= 2.0 seconds (2 sd)

15 m/s



If you don’t feel

so confident about

stringing them together,

you can do the units

conversion one step at a

time. That’s fine too.

The problem gave numbers

with 2 significant digits

to work with, so your

answer should have 2 sd.



The Emu takes 2.0 seconds to

reach the target - so the cage

needs to take 2.0 seconds to

reach the target as well.



NOTES



You know that

the Emu takes

2.0 s to arrive

at the target.



What time does the cage fall for?

The cage falls for 2.0 s.

What height should the crane be?

Will the cage be going faster than

25 m/s when it hits the ground?



00:02.00



So the cage needs to take

2.0 s to fall from the crane.



208   Chapter 6

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00:02.00



displacement, velocity, and acceleration



So I have the time figured out, but uh ... I still don’t know

how high the crane should be, or how fast the cage is going

when it hits the ground. Isn’t that the point?



Don’t be afraid

to start out doing

a question, even

if you’re not quite

sure what direction

it’s going to take.



Don’t worry - you’ve already made progress.

When you started out, you knew a couple of facts about the

Emu’s speed and the distance he covers - but nothing at all

about the cage or the crane platform.

Now we need to figure out how fast the cage is going when

it hits the ground after 2.0 s and the distance it falls in that

time.



BE the cage



Your job is to imagine that you’re the cage.

What do you feel at each of the points

in the picture? Which direction are you

moving in? Are you speeding up

or slowing down? Why are you

moving like this?



At Point 1:



Point 1 - Just

been pushed off

the platform



Point 2



At Point 2:

Point 3



At Point 3:

At Point 4:



Point 4 - Just

about to land

(but hasn’t hit

the ground yet)



Why:



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be the solution



BE the cage - SOLUTION



Your job is to imagine that you’re the cage.

What do you feel at each of the points

in the picture? Which direction are you

moving in? Are you speeding up

or slowing down? Why are you

moving like this??



Point 1 - Just

been pushed off

the platform



Point 2



At Point 1: A ‘special point’, as I’m suddenly going

from standing still to starting to move downwards.

At Point 2: Falling down faster than I was at

point 1.

At Point 3: Falling down even faster than I was at

point 2..

At Point 4: This is the fastest I’ll be going before

I hit the ground (I’ll be here after 2.0 seconds if

the height is right).

Why: Gravity’s accelerating me downwards.



Point 3



Point 4 - Just

about to land

(but hasn’t hit

the ground yet)



In this problem, we gave you headings to use,

but it’s always a good idea to make it clear

which part of the problem you’re answering

at each stage!



The cage accelerates as it falls

We’re going to talk about

the cage’s displacement

and velocity, as the

DIRECTION is starting

to become important

- the cage isn’t being

launched up into the air,

just dropped!



You’ve spotted that the cage accelerates as it falls.

Acceleration is the rate of change of velocity. You

can tell that the cage is accelerating because its velocity

is continually changing. It starts off with zero velocity,

then gets faster and faster until it hits the ground.

With that in mind, it’s on to working out the cage’s

velocity after 2.0 seconds and its displacement in

that time so that the Dingo knows whether the idea’s a

starter - and if so, how high to make the platform.



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You know that

something is

accelerating if

its velocity is

changing.



displacement, velocity, and acceleration



Hey ... what’s with this talk of

displacement and velocity? I was quite

happy with distance and speed.



Displacement and velocity will be

more useful to you in the long term.

As the cage is always falling in the same

direction - straight down - you could use either

distance and speed or displacement and velocity

to describe its motion.

But soon you’re going to be dealing with

situations where direction is crucially important,

and you must use vectors. As you practice using

displacement, velocity, and acceleration for the

cage, you’ll soon get comfortable with them,

which will stand you in good stead in the future.



Displacement, Velocity,

and Acceleration Up Close

Displacement is the ‘vector version’ of

distance and is represented by the letter x in

equations (or the letter s in some courses).

Velocity is rate of change of displacement

- the ‘vector version’ of speed. It is

represented by the letter v in equations.

Acceleration is rate of change of velocity,

represented by a, and doesn’t have a scalar

equivalent. If an object’s velocity is changing,

you need to know which direction the

velocity is changing in for the statement to

have meaning. Otherwise, you don’t know if

the object’s speeding up, slowing down, or

changing direction - which are all ways that

an object’s velocity can change.



‘ Vectorize’ your equation

You’ve already used the equation

distance to work out that it takes

speed =

time

the Emu 2.0 seconds to reach the target.

means

The ‘vector version’ of this equation is



‘change in’



displacement

x

, or v = t.

time

It’s fundamentally the same, except that it

involves velocity and displacement instead of

speed and distance.

velocity =



velocity



v=



x

t



change in

displacement



change in time



We’re using bold letters, like x and v, to represent

vectors and italic letters, like t, to represent scalars.



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displacement or velocity first?

So we need to work out the displacement of the

cage after 2.0 seconds. That doesn’t sound too bad.



Jim: We also need to work out what its velocity will be when it hits

the ground. If that’s more than 25 m/s, then the cage will shatter.

Joe: Why don’t we work out the velocity first? That way, if it turns

out that the cage is going too fast after it’s been falling for

2.0 seconds, we won’t have to bother working out the displacement

as well.

Frank: Sounds good. I’m all for spotting shortcuts!

Jim: Well, we’ve done something similar before with that cyclist

who rode everywhere at the same speed. Can’t we use the equation

x

v = t to work out the cage’s velocity

Frank: Yeah, let’s just use that equation! We want to know the

velocity, and that equation says “v =” on the left hand side. v for

velocity. It’s perfect!

Joe: Um, I’m not so sure. The cage doesn’t have the same velocity

all the time - it accelerates as it falls.

Jim: But we can still use that equation, right? If we work out the

displacement, we can divide it by the time to get the velocity.

Joe: I don’t think so. If the cage always had the same velocity, then,

fair enough, that would work. But the cage’s velocity is always

changing because it’s accelerating - it isn’t constant. We want to

know what its velocity is at the very end, as it hits the ground.

Frank: Oh ... and when it hits the ground, it’s only been traveling at

that velocity for a split second.

Jim: Yeah, as it gets closer to the ground its velocity increases, so

it covers more and more meters per second. If we divided the total

displacement by the total time, we’d get the cage’s average velocity.

Joe: But we need to know what the velocity is the instant it hits the

ground. An average velocity’s no good to us.

Frank: I guess we need to do something different ...



If you calculate the cage’s velocity first, you

won’t have to bother calculating its displacement

if it turns out that the cage will break.

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NEVER blindly stick

numbers into an

equation. Always ask

yourself “What does

this equation MEAN?”

NOTES

What time does the cage fall for?

The cage falls for 2.0 s.

What height should the crane be?

(Come back to this if necessary.)

Will the cage be going faster than

25 m/s when it hits the ground?

DO THIS NEXT!!



displacement, velocity, and acceleration



You want an instantaneous velocity,

not an average velocity

x

The equation v = t works fine if you have

something traveling at a constant velocity. But

the cage gets faster and faster as it falls - and you

want to know what its velocity is the instant it

hits the ground.



∆x

v

∆x



The best you can do with the equation is to work

out the cage’s average velocity, which is the

constant velocity it would need to travel with

to cover that displacement in that time. But since

the cage isn’t traveling with a constant velocity,

this value won’t help you out.



v



∆x



As its velocity increases, the

cage’s displacement is greater

in the same amount of time.



This vector represents the velocity of the

cage just before it hits the ground. The

length of the vector represents the size

of the velocity. Don’t be put off by it

appearing to go ‘into’ the target.



This is the

graph for

the cyclist in

chapter 4.



Plot of displacement vs time

for Alex’s late delivery



x

t



Displacement

(meters)



vavg =



1000

900

800

700



x



600

500



As it falls, the

cage’s velocity

increases.

The acceleration is the

rate at which the velocity

of the cage changes.



v



Strictly speaking, you used distance and

speed rather than displacement and

velocity, but the principle is the same.



x

You’ve previously used the equation v = t to work out the

average velocity of a cyclist who was slowed down by stop lights,

and it gave you the slope of a straight line between the start and

end points of his displacement-time graph. Using the slope of

his displacement-time graph at that point, you were also able to

work out his instantaneous velocity at any point.



400

300

200



How might you try to work out a value

for the instantaneous velocity of the

cage just before it hits the ground.



100

0



0



10



20



30



40



t



50



60



70



80



Time



(minutes)



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methods sometimes work when an equation won’t



So could we draw a displacement-time graph

for a falling thing, and calculate its slope at t = 2.0 s to get

its instantaneous velocity? Will that part still work?



You may be able to use the same method

even if you can’t use the same equation.

As the cage doesn’t fall with a constant velocity, the best you

x

can do with the equation v = t is work out its average

velocity - which isn’t what you want. You can’t reuse this

equation to work out the cage’s instantaneous velocity

because the context is different.

But you can use the same method even if you can’t

directly reuse the same equation. If you draw a

displacement-time graph for a falling thing and are

able to calculate its slope at t = 2.0 s, this will give you

the instantaneous velocity of the cage. As long as you

understand the physics, you can work out how to do a

problem even if you can’t directly use an equation you

already know.

Though you still need to design the experiment...



... but didn’t we already

design an experiment

like this?

Is the experimental setup you now

have in mind similar to what you drew

at the start - or is it different?

If it’s different, draw and label a diagram

of your new experimental setup - and

explain how you’ll use it

to make measurements and draw a

graph that shows you a value for the

displacement at any time.

If it’s the same as what you already

did, you can skip this Sharpen. :-)



Make sure you

include labels so

it’s clear what Electromagnet

everything is.



Ball-bearing



Timer



Clamp stand



Distance from

bottom of

ball-bearing

to top of

switch plate

(tape measure)



Switch plate

Don’t spend too much

time making your

diagram look pretty.



Use the clamp stand and the tape measure to set

the height of the ball-bearing. Time how long

it

takes to fall from that height using the timer,

electromagnet and switch plate. Use a range of

heights, from the smallest the timer can measure

to the height of the ceiling, and several heights

in between as well. And time each height two or

three times to reduce random errors.

The plot a graph with the time along the

horizontal axis and the distance up the vertical

axis. Draw a smooth line through the data points.

The graph lets you read off the time it’ll take

for the ball-bearing to fall any distance.

On your graphs, time should always

be along the horizontal axis.



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Understanding

the physics helps

you to work out

how to solve a

problem even if

you can’t directly

use an equation

you already know.



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