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Chapter 7. Equations of motion (part 1)

Chapter 7. Equations of motion (part 1)

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what’s the displacement?



How high should the crane be?

The Dingo wants to invite the Emu to his birthday party - but

the only way he’ll get him to stay still for long enough is by

catching him in a cage!



The Emu takes

2.0 s to arrive

at the target.



In chapter 6, you figured out that it takes the Emu 2.0 s to get

from the corner to the target on the road while running at his

constant speed.

You also figured out that the cage’s velocity after 2.0 s won’t

lead to it shattering on impact, by drawing its velocity v

time graph and working out the equation a = t .

But the Dingo wants to know how high to set the crane.

Which means that you now need to work out the cage’s

displacement after it’s been falling for 2.0 s.



00:02.00



What is the

displacement of

the cage after it’s

fallen for 2.0 s?



The falling cage’s displacement

- time graph is curved, so you

can’t extrapolate...

Maybe the

graph does

this ...



... or this ...



... or this ...



When you drew the displac

ement

- time graph for the cyclist,

you were

able to extrapolate it further

than the

measurements you’d origina

lly made.

Now you can extrapolate your

graph for

the ball bearing experiment.

Your current

set of measurements goes

up to 0.78 s

but you’re interested in what’s

going

on after it’s been falling for

2.0 s. We’ve

redrawn it to give you more

space.



x (m)



Graph of displacement vs.



... or this ...

... or maybe

even this ...



... or maybe it doesn’t

do any of these

things at all and does

something else instead!



You already drew a displacement - time

graph for a falling object, but you can’t

extrapolate it to read off the displacement

after 2.0 s because the graph is curved.



time



3.00

This TOTALLY stinks!!

Extrapolating from a straigh

t

line is fine, but how am I

supposed

to deal with curves when

there are

so many options?!



0.78



It’s only

meaningful to

extrapolate a

graph if its

points lie along

a straight line.



t (s)



It’s nearly impossible

to

extrapolate a curve accu

rately

It turns out that this method

isn’t so hot after

all. Drawing the displacement

- time graph

was fine, but this time, instead

of being a

straight line, it’s a curve.

With a straight line displacement

- time graph,

its easy to use a ruler to continu

e the straight

line as far as you need to.

But you can’t extrapolate from

a curved graph

in the same way, as it’s almost

impossible to

tell exactly how the curve will

continue.



If you can’t read the value for the

displacement after 2.0 s off your

graph, what can you do?



238   Chapter 7

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equations of motion (part 1)



OK, so we have to figure out the cage’s

displacement after it’s been falling for 2.0 s.

Are we absolutely sure we can’t just extrapolate

the displacement-time graph we drew before?



The limit of your

experiment in chapter 6

was dropping something

from 3.00 m, which

took 0.78 s.



Displacement (m)



3.00



0.78



Time (s)



Jim: It’s a curve, so we don’t really know what it’s going to do next. If

the last point we’d plotted was close to 2.0 s we could probably make an

educated guess, but not when we’re so far away.

Frank: But 0.78 s is only a little bit less than 2.0 s. We’d only need to

continue the graph for another 1.22 s - that’s hardly any time at all!

Jim: It’s a lot of time compared to what we already have. We’ve plotted

less than half the graph between t = 0.0 s and t = 2.0 s.

Joe: Maybe we could try working out an equation, like we did before

to get a value of the cage’s velocity from its velocity - time graph?



Velocity



Acceleration is

rate of change of

velocity with time.



∆v



v

t



a =



Time



v

t



Rearrange

equation to say

v =  something



= a t

= 9.8 × 2.0

= 20 m/s (2 sd)



You did this

in chapter 6.



Graphs and

equations are

both ways of

representing

reality.



Frank: But the velocity - time graph is a straight line.

Our displacement - time graph is a curved line!

Jim: Yeah, I dunno if it’s possible for a curved graph

to be represented by an equation.

Joe:: I’m sure it must be possible, if graphs and

equations are both ways of representing reality ...



Do you think it’s possible for a

curved graph to be represented

by an equation?



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graphs and equations



Graphs and equations both represent the real world

Graphs are a way of representing the real world visually.

Equations are a way of representing the real world

symbolically. They both allow you to predict what will

happen to a quantity when other things that affect it

change as well.



The graph and the

equation both represent

the same physical reality.



v Velocity - time



v

For example, a = t is the equation for the cage’s

velocity-time graph. The equation and the graph both

represent the same physical reality. The equation shows

you symbolically how the velocity, acceleration and time

interrelate when the acceleration is constant. If you know

values for two of the quantities in the equation, you can

use the equation to calculate the third by rearranging

the equation.



You can’t extrapolate this

graph, as it’s curved.



But if you can work

out an equation

that represents the

same thing, you can

solve the problem.



a=



v

t



t



x Displacement - time



?????

t



So, if you can work out the equation that

represents your displacement - time

graph, you’ll be able to use it to solve the

problem of how high the cage needs to be.



But I thought we said earlier that

we can’t form an equation using ∆x and ∆t

because our graph isn’t a straight line?



That’s right - we’re not going to use

x and t this time.

Originally, x and t helped with the concept of

finding the slope of a graph using the change

in x and t between two points. But as the slope

of this graph is continually changing, you’d have

to put the two points so close together it’d be

impossible to measure the changes!

Instead, you’ll use a different variable to

represent the displacement and time at each

point you’re interested in.

240   Chapter 7

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Use a different

variable to represent

each of the values

at the points you’re

interested in.



equations of motion (part 1)



You’re interested in the start and end points

We’re only really interested in two points in the cage’s motion - the start

(when it’s on the platform) and the end (when it hits the ground) - as we

want to calculate the cage’s displacement between these points.

In the equation you work out for your curved displacement - time

graph we’re going to use variables to represent every value we might be

interested in at these start and end points:

x0 is the displacement at the start (when t = 0).

v0 is the velocity at the start (when t = 0).

x is the displacement at the end.

v is the velocity at the end.

a is the acceleration (which you already know is constant).















The little ‘0’ is part of the variable

name, and is called a subscript.



There’s no point in having a0 and a,

as the value for the acceleration is

always the same.



We’ve drawn in the interesting start and end points

on your velocity - time graph.



velocity



You call this “v nought” if

you’re speaking out loud.

You can tell that v and v0 are

both displacements because they

use the letter v. But that they’re

different quantities because they

have different subscripts.



Use the same letter to

represent the same type

of thing, and subscripts

to say which is which.



b. Use the values on the graph to rewrite this

equation as an equation involving a, v0, v and t.



Graph of velocity vs time for a

falling object



End point.



You might

find it

helpful to

draw or

write on

the graph.

Start

point.



v



v0



This is how the

velocity of the

cage would

continue to change

if it hadn’t just

hit the ground!

0



t



c. Rearrange your equation so that it says

“v = something”.



time



a. Write down an equation you already know that

involves a, v and t.



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general equations



We’ve drawn in the start and end points we’re

interested in on your velocity - time graph.



velocity



b. Use the values on the graph to rewrite this as

an equation involving a, v0, v and t.



v - v0

=

t-0

v - v0

a =

t



a =



Graph of velocity vs time for a

falling object



v = v - v0



v



∆v

∆t



c. Rearrange your equation so that it says

“v = something”.



v - v0

t

v - v0

at =

t ×t

v0 + at = v - v0 + v0

a =



v0



t=t-0



0



t



time



a. Write down an equation you already know that

involves a, v and t.



Acceleration = Rate of change of velocity

a = tv



Now you have

v on its own.



v = v0 + at



Multiply both

sides by t.

Add v0 to

both sides.



Swap the

sides over.



Are we putting in letters, like ‘t’ for time, instead

of values, like 2.0 seconds, to make it more general?



You want your equation to be as general

as possible so you can use it elsewhere.

At the moment, you’re dealing with a falling cage. You

could stick in the numbers you already know (t = 2.0 s,

a = 9.8 m/s2) but then you’d end up with an equation

that you can only use once.

If you leave everything as letters for now and only

put the numbers in at the end, you’ll end up with a

general equation you can use to deal with falling

things, jet skis, racing cars ... anything that has a period

of constant acceleration.

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If your

equation is

general, you

can reuse

it in other

problems .



equations of motion (part 1)



Q:



Why did you choose particular variable names, like v (with

no subscripts) for the final velocity and v0 for the initial velocity?



A:



It’s a common convention to use x0 and v0 for the initial values

of displacement and velocity and x and v for the final values. It’s

what’s used in lots of textbooks, as well as the AP Physics B exam.



Q:



But the convention isn’t consistent! The initial velocity

is called v0, but the initial time doesn’t even have a symbol - we

just put in its value of 0.



Q:

A:



A graph and an equation can represent the same thing in real

life. In this problem, they both describe what happens to the velocity

of the falling cage as time passes.



Q:



OK. But why have I used letters in the equation when I

already worked out all the values of the things in the equation?!

I know what v, v0, a and t are for the falling cage!



A:



A:



Q:



Q:

A:



The convention assumes that everything you’re interested in

starts at t=0. The ‘0’ subscript in v0 stands for ‘at t=0’’, so you can

read v0 as “the velocity at t=0”. Similarly, t0 would stand for “the time

at t=0”. So there’s no need to bother with a t0 symbol, as you already

know that t=0 when t=0!



Do I have to use these letters? Before, I’ve used s instead

of x for displacement, and u instead of v0 for initial velocity. I’m

finding this confusing!



A:



The main thing is that you understand the physics concepts

that lie behind the equations. It doesn’t matter which set of letters you

use for that. It’s fine to show your work using the letters you’re more

familiar with that already make sense for you!



So what physics concepts are the most important here?



One reason is that you can reuse a general equation again and

again. If the crane is a different distance away from the corner, the

cage would fall for a different time. Your general equation, v = v0 + at,

will give you the value of v for any time. All you need to do is put in

the new numbers.

And the other reason for not putting in the values yet?



If you keep the equation general, you’ll be able to use it for

anything with constant acceleration, even if it isn’t 9.8 m/s2. All you

need to do is to put in the new numbers for your new problem.



Hey! We’re supposed to be figuring out

an equation for displacement, x. But the

equation we just worked out doesn’t have an x

in it, so how’s it gonna help?!?!



The equation shows how different variables

depend on one another. You can use it as a

stepping stone to get what you really want.

Your equation v = v0 + at shows you how the variables v, v0,

a and t depend on each other. But it doesn’t have an x in it, so

you can’t use it to directly calculate a value for the displacement,

However, as displacement is rate of change of velocity, the

displacement and the velocity must depend on each other. So

although you can’t use this equation directly, you’ll be able to

use it as a stepping stone towards calculating the displacement

of the cage after 2.0 s.



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velocity and displacement are related



You have an equation for the velocity but what about the displacement?

You’ve worked out the equation v = v0 + at , which

comes from the slope of your velocity - time graph. It

gives you an object’s velocity, v, after a certain amount

of time (if you know its initial velocity, v0, and its

acceleration, a).



The notebook keeps

track of where

you’re at so far.



Equation for the velocity

Used the slope of the velocity - time

graph to work out the acceleration,

then rearranged the equation.

Velocity

(m/s)



Graph of velocity vs time

for something with

constant acceleration



v = v - v0



v



v0



This equation

gives you the

velocity after a

certain amount

of time.



0



t=t-0



t



Time

(s)



v = v0 + at

But what we’re really interested in is the

displacement, x, after a certain amount of

time. If you have an equation for that, you

can say how far the cage will fall in 2.0 s.

The velocity equation might be useful later on,

as velocity and displacement must be related

somehow. But right now, you really need an

equation with an x in it to move forward ...



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For a falling thing, v0 = 0 m/s

downwards if you drop it from

a standing start

For a falling thing, a = 9.8 m/s2

at all times.



But the equation doesn’t

have x in it, which is what

you’re interested in!



How might you get an

equation that involves

the displacement?



equations of motion (part 1)

What about the average velocity?

Doesn’t that have something to do

with displacement and time?



The average velocity

is the same as the

constant velocity

you could have

gone at to cover

the displacement

between your start

and end points in

the same time



Displacement



Get the average velocity from the

total displacement and total time.

The average velocity of the cage between its

start and end points is given by the change

in its displacement divided by the change in

x

time, vavg = t

vavg is the average velocity - the same as the

constant velocity that an object would need

to travel with to cover that displacement in

that time.

As ∆x is the change in the displacement

between the start and end points, the

equation for the average velocity will have an

x in it - which is what you want to calculate!



x is the displacement

at the end point.



Graph of displacement vs time

for a falling object



a. Draw a line on your displacement - time

graph to represent the cage’s average velocity

between times 0 and t.

b. Use the graph to come up with an equation

for the average velocity, vavg , in terms of x0, x

and t.



x



x0



t



0



Time



You’re working out the

average velocity between

your start and end points.



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sharpen solution



Graph of displacement vs time

for a falling object



Displacement



a. Draw a line on your displacement - time

graph to represent the cage’s average velocity

between times 0 and t.



a. Average velocity is

the slope of this line.



b. Use the graph to come up with an equation

for the average velocity, vavg , in terms of x0, x

and t.



x



Total displacement

Total time

x - x0

= xt =

t-0



Average velocity =

x0



Time



t



0



vavg



You’re working out the

average velocity between

your start and end points.



vavg =



y

Equation for the velocit

locity - time

Used the slope of the ve

celeration,

graph to work out the ac

uation.

then rearranged the eq

Velocity

(m/s)



Graph of velocity vs time

for something with

constant acceleration



x - x0

t



Equation for the average velocity

Used the displacement - time graph to

work out the average velocity.

Displacement

(m)



Graph of displacement vs time

for something with

constant acceleration



x



v = v - v0



v



x0

0

v0



0



t

t=t-0



Time

(s)



v = v 0 + at

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t



vavg = x - x0

t



Time

(s)



An equation with x in it,

which is what I want!



equations of motion (part 1)



This stinks! Now we’ve got an equation with an x in it,

but it’s got the cage’s average velocity in it as well - and

we don’t know what that is!! So we can’t work out the value of

the displacement. How’s that supposed to help?!



That’s right - we don’t know the

value of the average velocity.

You’ve come a long way, and have two

equations from the graphs you drew:

x - x0

.

v = v0 + at and vavg =

t

The second of these equations has an x in it,

which is what you want - but it also has vavg,

the average velocity, in it.

Since you don’t know what the average

velocity is, you can’t use this equation to

calculate x and tell the Dingo how high to

put the crane platform right now. But you’re

definitely making progress ...



Wouldn't it be dreamy

if we could calculate the

value of the average velocity

a different way. But I know it’s

just a fantasy ...



x0 = 0 m



You want to

know x.

You know x0.



00:02.00



You know t.



t = 2.0 s

x=?



vavg =



x - x0

t



But you don’t

know vavg, so you

can’t figure out x.

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Equation for the velocity

- time

Used the slope of the velocity

tion,

substitute

for

unknown

variables

lera

graph to work out the acce

.

tion

then rearranged the equa

Velocity

(m/s)



Equation for the average velocity

Used

the

So we’ve

displacem

ent -got

timetwo

graphequations

to work out- that’s

gotta

beGraph

a ofgood

start.

the

average

velocity.

displacement vs time



Graph of velocity vs time

for something with

constant acceleration



Displacement

(m)



∆v = v - v0



v



for something with

constant



x



Time

Equatio0n for the velocityt

(s)

∆t = t - 0

time

city

Used the slope of the velo

tion,

graph to work out the accelera

.

tion

equa

the

then rearranged

0

v0



v = v + at



Velocity

(m/s)



x0

0



t



v = x-x

t



Time



(s)

Equation for the average velocity

0

Used avg

the displacement - time

graph to

work out the average velocity.

equation vswith

GraphAn

of displacement

time x in it,

Displacement

for something

which with

is what I want!

(m)



Graph of velocity vs time

for something with

constant acceleration



v



∆v = v - v0



constant acceleration

Jim: Let’s just see how they help us. I’m gonna put a question

mark by

the x, because we want to work that out. Then I’llx tick what we already

v

know

values for, and

cross

t

Timethe variables we don’t know ...

0



0



∆t = t - 0



(s)



v = v 0 + at



x0

0



t



vavg = x - x0

t



Time

(s)



An equation with x in it,

which is what I want!



Jim: ... hmmm, neither equation helps us. The one on the left is for the

velocity, v, which we’re not interested in. The one on the right has the

displacement, x, in it, which is what we want to work out ... but it also

has the average velocity, vavg, in it. And we don’t know what vavg is.

Joe: Is there another equation we can use?

Frank: What do you mean?

Joe: Our problem is vavg, right? We can’t just rearrange the equation

to say “x = something” and put in the values for the other variables

because we don’t have a value for vavg. But what if there was another

equation we could use to calculate the value of vavg?



Frank: I like your thinking. But we already worked out vavg the only way

we know how - from the slope of our displacement - time graph.

Jim: Hang on! What about our velocity - time graph? Maybe if we

look at that, we can eyeball a second equation for vavg.

Joe: You might be on to something there ... let’s try it!



248   Chapter 7

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If you don’t

know the value

for a variable

in your equation,

try to find

another equation

which includes

that variable.



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