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
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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?
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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 handdrawn 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
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v
a = 9.8 m/s2
equations of motion (part 2)
If your equation represents
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 nonzero 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
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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 nonzero 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
of your equation, v0t and ½at2, must add up to zero.
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.
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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
displacementtime 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.
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opposite directions opposite signs
Q:
I still don’t get how you can add two numbers together
and get zero as your answer.
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
direction instead?
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
what size your answer’s going to be at the back of your mind, then
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.
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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 displacementtime 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 accelerationtime 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
your equation.
Joe: Yeah, well, acceleration is rate of change of velocity. So the
value of the acceleration is the slope of the velocitytime graph.
The value of the acceleration is constant: 9.8 m/s2, so the slope of
the velocitytime 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
instead of up?! Like going downhill instead of uphill?
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
displacementtime graph.
Frank: Which lets us draw a displacementtime graph as well.
Jim: Great!
Don’t worry about numbers
here  the important thing
is the SHAPES of the graphs.
What might the velocitytime and
displacementtime graphs look like?
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negative and positive slope
The same principle applies to the
slope of a displacementtime 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
velocitytime 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 velocitytime
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 velocitytime
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.
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You already sketched graphs by
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)
Displacementtime
x
x
Displacementtime
Constant
gradient.
Increasing
gradient.
t
Velocity is constant
at all times, so always
has the same value.
So the velocitytime
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 velocitytime graph must be
constant and negative too.
v
Constant
value.
t
a
t
Velocitytime
v
Velocitytime
Increasing
value.
Zero
gradient.
t
Accelerationtime
a
Accelerationtime
Constant
value.
Zero
value.
Velocity is constant,
so acceleration is zero.
Constant
gradient.
t
t
Use the value of your accelerationtime graph to sketch the shape of the velocitytime
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)
worry about
displacementtime
the
Don’t worry
graph for the moment
about what
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 velocitytime
graph gives the value
of the acceleration.
Landing
The value of the
acceleration gives
the slope of the
velocity‑time graph.
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vectors have size and direction
Use the value of your accelerationtime graph to sketch the shape of the velocitytime
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 velocitytime graph has a constant slope of
9.8 m/s2 because the accelerationtime graph
has a constant value of 9.8 m/s2.
Time (s)
9.8
v
Acceleration (a) is
constantly 9.8 m/s2.
Velocitytime 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
heads of vectors are.
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.
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