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?
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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.
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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!
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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.