Answer key · Unit 2

Motion in a line.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Position and velocity

Where the cart is, with a sign, and how its position graph tells you how fast and which way it is going. A constant velocity is a straight line, and the velocity is its slope.

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The 6 steps and their answers

Step 1 of 6
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Position with a sign
Practice · Calculate
Question
The cart travelled towards the sensor and stopped at the end stop, 0.6 m from the start line on the sensor’s side. What is the cart’s position $x$ ⁠ now?

Correct answer
-0.6 m ± 0.01 m
Setup, choices and feedback
The Scene at this step
Starting velocity
-0.4 m/sfixed
Acceleration
0 m/s²fixed
Build the graphs
Position graphfixed
Other settings
Probe time/nodes/cart/props/probeTime = 4
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.01 m
Computed from the Scene at this step’s setting
Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.6 m ± 0.01 m
  Misconception: Gives a position on the sensor's side of the start line as a positive number, as if position were a distance.
  Feedback from the catalogue
  A position is not just a distance, it is also a side. The distance from the start line is 0.6 m, but positive is away from the sensor, so a position on the sensor’s side is negative: x = −0.6 m. Without the sign, 0.6 m could be on either side.
Feedback for a right answer
x = −0.6 m: six tenths of a metre from the start line, on the negative side. The number says how far, and the sign says which side.

Hint
Which side of the start line is the sensor on, and what is the sign of that side?

Worked solution
1. The zero point is the start line, and positive is away from the sensor.
2. The end stop is 0.6 m from the start line, on the sensor’s side: in the negative direction.
3. So x = −0.6 m.
If you wrote 0.6 m, you gave the distance without the side. x = 0.6 m is a different place altogether, six tenths of a metre past the start line, the other way.
Step 2 of 6
Open this step
Without a push
Practice · Predict, then look
Question
The cart will cross the start line at 0.3 m/s, away from the sensor. The fan is off and there is no friction. What will the position graph look like?

Correct answer
- Correct A straight line that rises by the same amount every second
Setup, choices and feedback
The Scene at this step
Starting velocity
0 m/s
Acceleration
0 m/s²fixed
Build the graphs
Position graphfixed
Choices
- Correct A straight line that rises by the same amount every second
- Wrong A curve that levels off, because the cart slows down and stops
  Misconception: Thinks a moving cart needs a push to keep moving: with the fan off it slows down and stops.
  Feedback from the catalogue
  With no push and no friction, nothing changes the cart’s velocity, so it carries on at the same speed in the same direction. A real cart stops in the end because friction and the air slow it down, not because “the push ran out”. The scene has no friction, and the position graph is a straight line.
- Wrong A curve that gets steeper and steeper
  Feedback
  A graph that gets steeper means a cart that is speeding up. With no push, nothing makes its velocity grow.
- Wrong A flat line
  Feedback
  A flat line is a cart standing still. Here the cart crosses the start line already moving.
Once the prediction is locked in, the Scene changes
Starting velocity
0.3 m/s
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With no force to change it, the velocity stays 0.3 m/s. The cart covers 0.3 m every second, and the position graph is a straight line. The orange marks on the track fall at equal spacings.

Worked solution
1. The fan is off and there is no friction, so nothing changes the velocity: it stays $0.3$ ⁠ metres per second.
2. At a constant velocity the position grows by the same step every second: 0.3 m, 0.6 m, 0.9 m, 1.2 m.
3. Points that rise in equal steps lie on a straight line, so the position graph is a line with slope $0.3$ ⁠.
If you expected the cart to slow down, that is what a real cart does, because of friction and the air. Without them, a moving cart keeps moving.
Step 3 of 6
Open this step
Velocity is the slope
Practice · Calculate
Question
The cart crosses the start line at 0.35 m/s, and the fan is off. What is its position after three seconds?

Correct answer
1.05 m ± 0.02 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.35 m/sfixed
Acceleration
0 m/s²fixed
Build the graphs
Position graphfixed
Other settings
Probe time/nodes/cart/props/probeTime = 3
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.02 m
Computed from the Scene at this step’s setting
Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$x = 0 + 0.35 \times 3 = 1.05$ ⁠ metres. You can also read it off the tape on the track as the clock passes 3 s.

Hint
Each second the cart covers 0.35 m. Use $x = x_0 + vt$ ⁠.

Worked solution
1. The givens: $x_0 = 0$ ⁠ (the clock starts at the start line), v = 0.35 m/s, t = 3 s.
2. The velocity is constant, so $x = x_0 + vt$ ⁠.
3. Substitute: $x = 0 + 0.35 \times 3 = 1.05$ ⁠.
4. The cart is 1.05 m past the start line.
If you got about $0.12$ ⁠, you divided the velocity by the time. Check the units: $\mathrm{m/s} \times \mathrm{s} = \mathrm{m}$ ⁠, while $\dfrac{\mathrm{m/s}}{\mathrm{s}}$ ⁠ is not metres.
Step 4 of 6
Open this step
Negative velocity
Practice · Calculate
Question
The position graph falls in a straight line: from $0$ ⁠ at t = 0 to −0.4 m at t = 4 s. What is the cart’s velocity?

Correct answer
-0.1 m/s ± 0.01 m/s
Setup, choices and feedback
The Scene at this step
Starting velocity
-0.1 m/shidden
Acceleration
0 m/s²fixed
Build the graphs
Position graphfixed
Accepted within
± 0.01 m/s
Computed from the Scene at this step’s setting
Velocity at probeTime (the last velocity before an end stop)/nodes/cart/outputs/velocityAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.1 m/s ± 0.01 m/s
  Misconception: Drops the minus sign of a velocity or an acceleration that points back towards the sensor.
  Feedback from the catalogue
  Velocity and acceleration have a direction, and the sign is the direction. A cart coming back to the sensor has a negative velocity, and a fan pushing towards the sensor gives a negative acceleration. 0.1 m/s and −0.1 m/s are the same speed and two opposite motions.
- -0.4 m/s ± 0.01 m/s
  Misconception: Reads the height of a graph where its slope is asked for: the position as the velocity, or the velocity as the acceleration.
  Feedback from the catalogue
  The height of the position graph is where the cart is; the velocity is how fast that height changes, the slope. The same goes for the velocity graph: its height is the velocity, and its slope is the acceleration. At the top of the position graph’s arch the cart is farthest away, but the tangent there is flat and the velocity is zero.
Feedback for a right answer
$v = \dfrac{-0.4 - 0}{4} = -0.1$ ⁠ metres per second. The graph falls, so the velocity is negative: the cart is travelling towards the sensor at 0.1 m/s.

Hint
The velocity is the slope: the change in position divided by the time that passed. Mind the sign.

Worked solution
1. From the graph: at the start $x = 0$ ⁠, and after 4 s $x = -0.4$ ⁠ metres.
2. The velocity is the slope: $v = \dfrac{\Delta x}{\Delta t}$ ⁠.
3. Substitute: $v = \dfrac{-0.4 - 0}{4 - 0} = -0.1$ ⁠.
4. v = −0.1 m/s: a speed of 0.1 m/s, towards the sensor.
If you wrote 0.1 m/s, you lost the direction. If you wrote −0.4, you gave the height of the graph at the end, the position, not its slope.
Step 5 of 6
Open this step
The equation of uniform motion
Practice · Choose
Question
A runner on a straight track is at x = 20 m when timing starts, and runs towards the zero point at a steady 4 m/s. Which equation gives her position, $x$ ⁠ in metres and $t$ ⁠ in seconds?

Correct answer
- Correct $x = 20 - 4t$ ⁠
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/s
Acceleration
0 m/s²fixed
Build the graphs
Position graph
Choices
Shown in a random order.
- Correct $x = 20 - 4t$ ⁠
- Wrong $x = 20 + 4t$ ⁠
  Misconception: Drops the minus sign of a velocity or an acceleration that points back towards the sensor.
  Feedback from the catalogue
  Velocity and acceleration have a direction, and the sign is the direction. A cart coming back to the sensor has a negative velocity, and a fan pushing towards the sensor gives a negative acceleration. 0.1 m/s and −0.1 m/s are the same speed and two opposite motions.
- Wrong $x = -4t$ ⁠
  Feedback
  This equation gives $x = 0$ ⁠ at $t = 0$ ⁠, but the runner started at $20$ ⁠ metres. The starting position $x_0$ ⁠ is missing.
- Wrong $x = 4 - 20t$ ⁠
  Feedback
  The numbers have swapped places. In $x = x_0 + vt$ ⁠ the number standing alone is the starting position, and the number multiplying $t$ ⁠ is the velocity.
Feedback for a right answer
$x_0 = 20$ ⁠ and v = −4 m/s, because she runs in the negative direction. After $5$ ⁠ seconds $x = 20 - 20 = 0$ ⁠: she is at the zero point.

Hint
Write $x = x_0 + vt$ ⁠. What is $x_0$ ⁠, and what is the sign of $v$ ⁠ when she is heading for zero?

Worked solution
1. Motion at a constant velocity: $x = x_0 + vt$ ⁠.
2. The position at $t = 0$ ⁠: $x_0 = 20$ ⁠.
3. She runs towards zero, so her position gets smaller: v = −4 m/s.
4. So $x = 20 - 4t$ ⁠. Check: after one second $x = 16$ ⁠, four metres closer to zero.
If you chose $x = 20 + 4t$ ⁠, your runner is moving away from zero: the velocity has to be negative.
Step 6 of 6
Open this step
Summary
No practice question in this step.
The Scene at this step
Starting velocity
0.6 m/s
Acceleration
0 m/s²fixed
Build the graphs
Position graph

Lesson 2 of 4 · About 25 minutes

Acceleration

When the fan pushes, the velocity changes by the same amount every second. What acceleration is, how to work out a velocity with it, and why a negative acceleration does not always mean slowing down.

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The 6 steps and their answers

Step 1 of 6
Open this step
A growing velocity
Practice · Calculate
Question
The cart sets off from rest, and the fan gives it an acceleration of 0.2 m/s². What is its velocity after three seconds?

Correct answer
0.6 m/s ± 0.02 m/s
Setup, choices and feedback
The Scene at this step
Starting velocity
0 m/sfixed
Acceleration
0.2 m/s²fixed
Build the graphs
Position graphfixed
Other settings
Probe time/nodes/cart/props/probeTime = 3
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.02 m/s
Computed from the Scene at this step’s setting
Velocity at probeTime (the last velocity before an end stop)/nodes/cart/outputs/velocityAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.2 m/s ± 0.02 m/s
  Misconception: Gives the acceleration as the velocity, as if a = 0.2 m/s² meant moving at 0.2 m/s.
  Feedback from the catalogue
  The acceleration is not the velocity but how much the velocity changes each second. An acceleration of 0.2 m/s² adds 0.2 m/s every second, so three seconds from rest the velocity is $3 \times 0.2 = 0.6$ ⁠ metres per second.
Feedback for a right answer
$v = at = 0.2 \times 3 = 0.6$ ⁠ metres per second: 0.2 m/s in each of the three seconds.

Hint
Each second the velocity grows by 0.2 m/s.

Worked solution
1. The givens: $v_0 = 0$ ⁠ (from rest), a = 0.2 m/s², t = 3 s.
2. The acceleration is the change in velocity each second, so $v = v_0 + at$ ⁠.
3. Substitute: $v = 0 + 0.2 \times 3 = 0.6$ ⁠.
4. After three seconds the cart is moving at 0.6 m/s.
If you wrote 0.2 m/s, you gave the acceleration as though it were the velocity. That is the velocity after just one second.
Step 2 of 6
Open this step
A cart that is already moving
Practice · Calculate
Question
The cart crosses the start line at 0.2 m/s, and the fan gives it an acceleration of 0.15 m/s² in the same direction. What is its velocity after two seconds?

Correct answer
0.5 m/s ± 0.02 m/s
Setup, choices and feedback
The Scene at this step
Starting velocity
0.2 m/sfixed
Acceleration
0.15 m/s²fixed
Build the graphs
Position graphfixed
Other settings
Probe time/nodes/cart/props/probeTime = 2
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.02 m/s
Computed from the Scene at this step’s setting
Velocity at probeTime (the last velocity before an end stop)/nodes/cart/outputs/velocityAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.3 m/s ± 0.02 m/s
  Misconception: Uses v = at for a cart that was already moving, leaving out its starting velocity.
  Feedback from the catalogue
  $v = at$ ⁠ only holds from rest. A cart that is already moving starts from its own velocity, and the acceleration adds to it: $v = v_0 + at$ ⁠. Without $v_0$ ⁠ you get only the change in velocity, not the velocity itself.
Feedback for a right answer
$v = 0.2 + 0.15 \times 2 = 0.5$ ⁠ metres per second. The starting velocity plus what was added in two seconds.

Hint
$v = v_0 + at$ ⁠: add what the acceleration added to the velocity the cart already had.

Worked solution
1. The givens: v₀ = 0.2 m/s, a = 0.15 m/s², t = 2 s.
2. With constant acceleration $v = v_0 + at$ ⁠.
3. Substitute: $v = 0.2 + 0.15 \times 2 = 0.2 + 0.3 = 0.5$ ⁠.
4. The velocity is 0.5 m/s.
If you got 0.3 m/s, you worked out only the change in velocity, $at$ ⁠, and forgot that the cart was already moving at 0.2 m/s.
Step 3 of 6
Open this step
A push against the motion
Practice · Predict, then look
Question
The cart is moving at 0.6 m/s away from the sensor. The fan will come on and push it towards the sensor, with an acceleration of −0.3 m/s², for the whole run. What will the cart do?

Correct answer
- Correct Slow down, stop for an instant and come back towards the sensor
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/sfixed
Acceleration
0 m/s²fixed
Build the graphs
All three graphs
Choices
- Correct Slow down, stop for an instant and come back towards the sensor
- Wrong Slow down, stop and stay there
  Misconception: Thinks a push against the motion brings the cart to rest and it then stays there.
  Feedback from the catalogue
  The backward push does not vanish when the cart stops. At the moment it stops the velocity is zero, but the acceleration is still −0.3 m/s², so an instant later the velocity is already negative: the cart comes back. It would stay put only if the fan switched off at exactly the moment it stopped.
- Wrong Keep slowing down, but never quite stop
  Feedback
  The velocity drops by $0.3$ ⁠ every second, the same amount each time, not by a share of what is left. After two seconds there is none left.
- Wrong Carry on at the same velocity
  Feedback
  A non-zero acceleration means the velocity changes. Here it drops by $0.3$ ⁠ metres per second every second.
Once the prediction is locked in, the Scene changes
Acceleration
-0.3 m/s²fixed
The Scene then shows: 2 s · When the cart stops and turns round (NaN if it does not within the run)/nodes/cart/outputs/turnTime
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The cart stops after two seconds, and the velocity graph crosses zero there without stopping on it. The fan keeps pushing, and the velocity keeps falling below zero: the cart comes back.

Worked solution
1. The givens: v₀ = 0.6 m/s, a = −0.3 m/s².
2. The velocity: $v = 0.6 - 0.3t$ ⁠.
3. The cart stops when $v = 0$ ⁠: $0.6 - 0.3t = 0$ ⁠, so $t = 2$ ⁠ seconds.
4. After that, at $t = 3$ ⁠ say, $v = 0.6 - 0.9 = -0.3$ ⁠: the velocity is negative, and the cart is travelling towards the sensor.
For the cart to stop and stay, the push would have to vanish at the moment it stops. Here the push goes on, and so does the change in velocity.
Step 4 of 6
Open this step
Negative acceleration, and a cart that speeds up
Practice · Predict, then look
Question
The cart is rolling towards the sensor at 0.1 m/s, that is v = −0.1 m/s. The fan will come on and push towards the sensor too, with an acceleration of −0.1 m/s². What will happen to the cart?

Correct answer
- Correct It will speed up
Setup, choices and feedback
The Scene at this step
Starting velocity
-0.1 m/sfixed
Acceleration
0 m/s²fixed
Build the graphs
All three graphs
Other settings
Probe time/nodes/cart/props/probeTime = 2
Choices
- Correct It will speed up
- Wrong It will slow down, because the acceleration is negative
  Misconception: Thinks a negative acceleration always means slowing down.
  Feedback from the catalogue
  The sign of the acceleration tells you which way the velocity is changing, not whether the cart is slowing down. A cart coming back to the sensor (negative velocity) and pushed towards the sensor (negative acceleration) speeds up: its velocity becomes ever more negative. A cart slows down when its velocity and acceleration have opposite signs, and speeds up when they have the same sign.
- Wrong It will carry on at the same speed
  Feedback
  The fan is pushing, so the velocity changes by $0.1$ ⁠ metres per second every second. The only question is which way.
Once the prediction is locked in, the Scene changes
Acceleration
-0.1 m/s²fixed
The Scene then shows: -0.3 m/s · Velocity at probeTime (the last velocity before an end stop)/nodes/cart/outputs/velocityAt
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The velocity and the acceleration have the same sign, so the cart speeds up. After two seconds v = −0.3 m/s: three times as fast, towards the sensor.

Worked solution
1. The givens: v₀ = −0.1 m/s, a = −0.1 m/s².
2. $v = v_0 + at = -0.1 - 0.1t$ ⁠.
3. After two seconds: $v = -0.1 - 0.2 = -0.3$ ⁠ metres per second. The speed has grown from 0.1 m/s to 0.3 m/s.
4. The cart speeds up: its velocity and acceleration have the same sign.
If you chose slowing down, you took a negative acceleration to mean slowing down every time. The sign of an acceleration is a direction, and it tells you what happens to the velocity only compared with the velocity’s own sign.
Step 5 of 6
Open this step
At the turning point
Practice · Choose
Question
In this run the cart stops for an instant after two seconds, and there it starts to come back. What is its acceleration at that moment?

Correct answer
- Correct −0.3 m/s², as at every other moment of the run
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/sfixed
Acceleration
-0.3 m/s²fixed
Build the graphs
All three graphs
Choices
Shown in a random order.
- Correct −0.3 m/s², as at every other moment of the run
- Wrong Zero, because the cart is standing still
  Misconception: Thinks that at the instant the velocity is zero, the acceleration is zero too.
  Feedback from the catalogue
  Zero velocity does not mean zero acceleration. The acceleration is how much the velocity changes, and at the turning point it keeps changing: an instant before it was positive, an instant after it is negative. The fan has not stopped pushing, so the acceleration stays −0.3 m/s². If it were zero, the cart would stay where it is.
- Wrong +0.3 m/s², because the cart is starting to come back
  Feedback
  A positive acceleration would push the cart away from the sensor. The cart comes back towards the sensor, so its velocity is becoming negative: the acceleration is negative.
Feedback for a right answer
The acceleration graph is a flat line, at the turning point too. The velocity is zero for just one instant, and it is still changing: from positive to negative.

Hint
Look at the acceleration graph at two seconds. Has the fan switched off?

Worked solution
1. The fan pushes back the whole time, so the acceleration is constant: a = −0.3 m/s².
2. Half a second before the turn $v = 0.6 - 0.3 \times 1.5 = 0.15$ ⁠, and half a second after it $v = 0.6 - 0.3 \times 2.5 = -0.15$ ⁠.
3. The velocity fell by $0.3$ ⁠ in the one second that passes through the turn, just as in every other second.
4. So at the turning point the acceleration is −0.3 m/s².
If you chose zero, you mixed up zero velocity and zero acceleration. If the acceleration were zero at that moment, the velocity would stay zero and the cart would not come back.
Step 6 of 6
Open this step
Summary
No practice question in this step.
The Scene at this step
Starting velocity
0.6 m/s
Acceleration
-0.3 m/s²
Build the graphs
All three graphs

Lesson 3 of 4 · About 25 minutes

Reading motion graphs

Three graphs of one trip. What the height says and what the slope says, how to read an acceleration off the velocity graph, and why the area under the velocity graph is the displacement.

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The 7 steps and their answers

Step 1 of 7
Open this step
The graph is not a picture of the path
Practice · Choose
Question
The position graph of this run is an arch: it rises, peaks after two seconds and falls back to zero after four. What did the cart do?

Correct answer
- Correct Moved away from the sensor, stopped for an instant and came back to the start line
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/sfixed
Acceleration
-0.3 m/s²fixed
Build the graphs
Position graphfixed
Choices
Shown in a random order.
- Correct Moved away from the sensor, stopped for an instant and came back to the start line
- Wrong Went up a hill and down the other side
  Misconception: Reads a position–time graph as a picture of the path, so an arch means the cart went up and down a hill.
  Feedback from the catalogue
  The graph is not a picture of the path. The track is straight and level, and the cart only ever moves along it. The graph’s vertical axis is the position on the track and its horizontal axis is time, so an arch means: it moved away, stopped and came back.
- Wrong Was going fastest after two seconds, at the top of the graph
  Misconception: Reads the height of a graph where its slope is asked for: the position as the velocity, or the velocity as the acceleration.
  Feedback from the catalogue
  The height of the position graph is where the cart is; the velocity is how fast that height changes, the slope. The same goes for the velocity graph: its height is the velocity, and its slope is the acceleration. At the top of the position graph’s arch the cart is farthest away, but the tangent there is flat and the velocity is zero.
Feedback for a right answer
Up to the peak the position grows: the cart is moving away. At the peak the tangent is flat and the velocity is zero. After it the position shrinks: the cart comes back, until it is on the start line again.

Hint
The graph’s vertical axis is the position on the track. What happens to the position up to the peak, and after it?

Worked solution
1. The height of the position graph is where the cart is, and its slope is the velocity.
2. From $0$ ⁠ to $2$ ⁠ seconds the graph rises: the position grows and the cart moves away from the sensor. The slope keeps shrinking, so it is slowing down.
3. At the peak the slope is zero: the cart stands still for an instant.
4. From $2$ ⁠ to $4$ ⁠ seconds the graph falls to $0$ ⁠: the cart comes back, as far as the start line.
A hill is a picture of a path, but the track is straight. And at the peak the cart is not fastest but farthest away: there it is not moving at all.
Step 2 of 7
Open this step
The slope of the velocity graph
Practice · Calculate
Question
The velocity graph falls in a straight line from 0.5 m/s at t = 0 to −0.5 m/s at t = 4 s. What is the cart’s acceleration?

Correct answer
-0.25 m/s² ± 0.01 m/s²
Setup, choices and feedback
The Scene at this step
Starting velocity
0.5 m/shidden
Acceleration
-0.25 m/s²hidden
Build the graphs
All three graphsfixed
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.01 m/s²
Computed from the Scene at this step’s setting
Acceleration/nodes/cart/props/acceleration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.25 m/s² ± 0.01 m/s²
  Misconception: Drops the minus sign of a velocity or an acceleration that points back towards the sensor.
  Feedback from the catalogue
  Velocity and acceleration have a direction, and the sign is the direction. A cart coming back to the sensor has a negative velocity, and a fan pushing towards the sensor gives a negative acceleration. 0.1 m/s and −0.1 m/s are the same speed and two opposite motions.
- 0.5 m/s² ± 0.01 m/s²
  Misconception: Reads the height of a graph where its slope is asked for: the position as the velocity, or the velocity as the acceleration.
  Feedback from the catalogue
  The height of the position graph is where the cart is; the velocity is how fast that height changes, the slope. The same goes for the velocity graph: its height is the velocity, and its slope is the acceleration. At the top of the position graph’s arch the cart is farthest away, but the tangent there is flat and the velocity is zero.
Feedback for a right answer
$a = \dfrac{-0.5 - 0.5}{4} = -0.25$ ⁠ metres per second squared. The acceleration graph below shows a flat line halfway between $0$ ⁠ and −0.5, in the same place.

Hint
The acceleration is the slope of the velocity graph: the change in velocity, end minus start, divided by the time.

Worked solution
1. From the graph: v₀ = 0.5 m/s, and after 4 s v = −0.5 m/s.
2. The acceleration is the slope: $a = \dfrac{\Delta v}{\Delta t}$ ⁠.
3. Substitute: $a = \dfrac{-0.5 - 0.5}{4 - 0} = \dfrac{-1}{4} = -0.25$ ⁠.
4. a = −0.25 m/s².
If you got 0.25, you swapped the order or dropped the sign: the velocity fell, so the acceleration is negative. If you wrote 0.5, you read the height of the velocity graph at the start, not its slope.
Step 3 of 7
Open this step
The area under the velocity graph
Practice · Calculate
Question
The cart’s velocity rises in a straight line from 0.1 m/s to 0.5 m/s in four seconds. What is its displacement in those four seconds?

Correct answer
1.2 m ± 0.02 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.1 m/shidden
Acceleration
0.1 m/s²hidden
Build the graphs
All three graphs
Other settings
Probe time/nodes/cart/props/probeTime = 4
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.02 m
Computed from the Scene at this step’s setting
Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.4 m ± 0.02 m
  Misconception: Multiplies one velocity, the starting one or the final one, by the time, though the velocity changed during the run.
  Feedback from the catalogue
  $x = vt$ ⁠ works only when the velocity is constant. When it changes, the displacement is the area under the velocity graph: a rectangle for the starting velocity plus a triangle for what was added. The starting velocity alone gives too little, and the final velocity alone too much. With constant acceleration you can use their average.
- 2 m ± 0.02 m
  Misconception: Multiplies one velocity, the starting one or the final one, by the time, though the velocity changed during the run.
  Feedback from the catalogue
  $x = vt$ ⁠ works only when the velocity is constant. When it changes, the displacement is the area under the velocity graph: a rectangle for the starting velocity plus a triangle for what was added. The starting velocity alone gives too little, and the final velocity alone too much. With constant acceleration you can use their average.
Feedback for a right answer
A rectangle of $0.1 \times 4 = 0.4$ ⁠ with a triangle of $\tfrac12 \times 4 \times 0.4 = 0.8$ ⁠ on top: 1.2 m in all. At the end of the run the cart is 1.2 m past the start line.

Hint
The displacement is the area under the velocity graph: a rectangle $0.1$ ⁠ high with a triangle on top.

Worked solution
1. The givens: the velocity rises from 0.1 m/s to 0.5 m/s in 4 s.
2. The displacement is the area under the velocity graph, a trapezium: a rectangle with a triangle on top.
3. The rectangle: $0.1 \times 4 = 0.4$ ⁠. The triangle: base $4$ ⁠, height $0.5 - 0.1 = 0.4$ ⁠, area $\tfrac12 \times 4 \times 0.4 = 0.8$ ⁠.
4. Together: $0.4 + 0.8 = 1.2$ ⁠, so 1.2 m. The same comes from the average velocity: $\dfrac{0.1 + 0.5}{2} \times 4 = 1.2$ ⁠.
If you got 0.4 m, you used only the starting velocity, the rectangle. If you got 2 m, you multiplied the final velocity by the time, but the cart moved that fast only at the very end.
Step 4 of 7
Open this step
Distance and displacement
Practice · Calculate
Question
In this run, v₀ = 0.6 m/s and a = −0.3 m/s², the cart is back on the start line at the end of the four seconds, and its displacement is zero. What distance did it travel?

Correct answer
1.2 m ± 0.02 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/sfixed
Acceleration
-0.3 m/s²fixed
Build the graphs
Area under velocity
Other settings
Probe time/nodes/cart/props/probeTime = 4
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.02 m
Computed from the Scene at this step’s setting
Distance travelled by probeTime, counting both ways after a turn/nodes/cart/outputs/distanceAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 m ± 0.02 m
  Misconception: Mixes up the distance travelled and the displacement, giving one for the other after the cart has turned.
  Feedback from the catalogue
  The displacement is the difference between where the cart ends and where it started, and area below the time axis counts against it. The distance travelled counts every metre the cart covered, out and back, so it is never negative. When the cart does not turn round they are equal; after a turn the distance is bigger than the size of the displacement.
Feedback for a right answer
0.6 m out and 0.6 m back: a distance of 1.2 m, though the displacement is zero.

Hint
Find the area of each of the two triangles under the velocity graph, and count both as positive.

Worked solution
1. The cart stops when $0.6 - 0.3t = 0$ ⁠, that is after $2$ ⁠ seconds.
2. The triangle above the axis, from $0$ ⁠ to $2$ ⁠ seconds: $\tfrac12 \times 2 \times 0.6 = 0.6$ ⁠. That is the trip out, 0.6 m.
3. The triangle below the axis, from $2$ ⁠ to $4$ ⁠ seconds: the velocity falls to −0.6 m/s, and its area is $0.6$ ⁠. That is the trip back.
4. The distance counts both: $0.6 + 0.6 = 1.2$ ⁠, so 1.2 m. The displacement subtracts them: $0.6 - 0.6 = 0$ ⁠.
If you wrote $0$ ⁠, you gave the displacement. A distance never cancels out, because every metre the cart covered counts.
Step 5 of 7
Open this step
Average velocity
Practice · Calculate
Question
The cart crosses the start line at 0.6 m/s with an acceleration of −0.2 m/s². It stops after three seconds 0.9 m from the start line, and comes back to 0.8 m by the end of the four seconds. What is its average velocity over the four seconds?

Correct answer
0.2 m/s ± 0.01 m/s
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/sfixed
Acceleration
-0.2 m/s²fixed
Build the graphs
All three graphs
Accepted within
± 0.01 m/s
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.25 m/s ± 0.01 m/s
  Misconception: Mixes up the distance travelled and the displacement, giving one for the other after the cart has turned.
  Feedback from the catalogue
  The displacement is the difference between where the cart ends and where it started, and area below the time axis counts against it. The distance travelled counts every metre the cart covered, out and back, so it is never negative. When the cart does not turn round they are equal; after a turn the distance is bigger than the size of the displacement.
Feedback for a right answer
$\bar v = \dfrac{0.8 - 0}{4} = 0.2$ ⁠ metres per second. That is also the average of the starting and final velocities: $\dfrac{0.6 + (-0.2)}{2} = 0.2$ ⁠.

Hint
Average velocity is the displacement, the final position minus the starting one, divided by the time.

Worked solution
1. The givens: at the start $x = 0$ ⁠, after 4 s $x = 0.8$ ⁠ metres.
2. The average velocity: $\bar v = \dfrac{\Delta x}{\Delta t}$ ⁠.
3. Substitute: $\bar v = \dfrac{0.8 - 0}{4} = 0.2$ ⁠, so 0.2 m/s.
4. Check: the final velocity is $0.6 - 0.2 \times 4 = -0.2$ ⁠, and the average of $0.6$ ⁠ and $-0.2$ ⁠ is $0.2$ ⁠. With constant acceleration the two ways agree.
If you got 0.25 m/s, you divided the distance, $0.9 + 0.1 = 1.0$ ⁠ metres, by the time. That is the average speed, not the average velocity.
Step 6 of 7
Open this step
Out and back
Practice · Reach the target
Question
Find a run in which the cart is back on the start line exactly at the end of the four seconds, having travelled a distance of 0.8 m. Change Starting velocity and Acceleration, and press Check.

Goal
- 0 m ± 0.02 m
  At the start of the step: 0.8 m
- 0.8 m ± 0.02 m
  At the start of the step: 1 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/s
Acceleration
-0.2 m/s²
Build the graphs
Area under velocity
Other settings
Probe time/nodes/cart/props/probeTime = 4
Read from the Scene at
Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
Read from the Scene at
Distance travelled by probeTime, counting both ways after a turn/nodes/cart/outputs/distanceAt
One setting that reaches it
Starting velocity
0.4 m/s
Acceleration
-0.2 m/s²
Feedback for a right answer
Two equal triangles under the velocity graph, one above the axis and one below: the displacement is zero and the distance is twice the area of each.

Feedback for a wrong answer
Not yet. If the cart does not get back to the start line, it has to turn at exactly two seconds. If the distance is off, change both in the same ratio.

Hint
A run that comes back to the start line turns halfway, after two seconds, and there it is half the distance away: 0.4 m.

Worked solution
1. To be back on the start line after $4$ ⁠ seconds, the cart turns halfway, after $2$ ⁠ seconds: $v_0 + 2a = 0$ ⁠, so $v_0 = -2a$ ⁠.
2. The distance is twice the distance to the turn, so the turn is 0.4 m from the start line. That is the first triangle: $\tfrac12 \times 2 \times v_0 = v_0$ ⁠, so $v_0 = 0.4$ ⁠.
3. That gives $a = -\dfrac{v_0}{2} = -0.2$ ⁠. Set Starting velocity to 0.4 m/s and Acceleration to −0.2 m/s².
4. Check: $x = 0.4 \times 4 - \tfrac12 \times 0.2 \times 16 = 1.6 - 1.6 = 0$ ⁠.
−0.4 m/s with +0.2 m/s² works too: the same run the other way, towards the sensor and back. 0.6 m/s with −0.3 m/s² comes back to the line, but travels 1.2 m.
Step 7 of 7
Open this step
Summary
No practice question in this step.
The Scene at this step
Starting velocity
0.6 m/sdefault
Acceleration
-0.3 m/s²default
Build the graphs
Position graphdefault

Lesson 4 of 4 · About 25 minutes

The equations of constant acceleration

From the area under the velocity graph to three equations that solve any motion with constant acceleration: where the object is, when it stops, and why twice the speed means four times the braking distance.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Position from the area
Practice · Calculate
Question
The cart crosses the start line at 0.4 m/s, with an acceleration of −0.1 m/s². What is its position after three seconds?

Correct answer
0.75 m ± 0.02 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.4 m/sfixed
Acceleration
-0.1 m/s²fixed
Build the graphs
All three graphsfixed
Other settings
Probe time/nodes/cart/props/probeTime = 3
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.02 m
Computed from the Scene at this step’s setting
Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.3 m ± 0.02 m
  Misconception: Drops the ½ (or the 2) in a constant-acceleration equation: x = v₀t + at², or a stopping distance of v₀²/a.
  Feedback from the catalogue
  The half comes from the triangle under the velocity graph: base $t$ ⁠, height $at$ ⁠, and area half their product, $\tfrac12 at^2$ ⁠. In the same way $v^2 = v_0^2 + 2a\Delta x$ ⁠ gives a stopping distance of $\dfrac{v_0^2}{2|a|}$ ⁠, not $\dfrac{v_0^2}{|a|}$ ⁠. Without the half the answer comes out twice as big.
- 1.2 m ± 0.02 m
  Misconception: Multiplies one velocity, the starting one or the final one, by the time, though the velocity changed during the run.
  Feedback from the catalogue
  $x = vt$ ⁠ works only when the velocity is constant. When it changes, the displacement is the area under the velocity graph: a rectangle for the starting velocity plus a triangle for what was added. The starting velocity alone gives too little, and the final velocity alone too much. With constant acceleration you can use their average.
Feedback for a right answer
$x = 0.4 \times 3 + \tfrac12 \times (-0.1) \times 9 = 1.2 - 0.45 = 0.75$ ⁠ metres. Without the push the cart would have reached 1.2 m; the negative acceleration took away 0.45 m.

Hint
$x = x_0 + v_0 t + \tfrac12 at^2$ ⁠, with $x_0 = 0$ ⁠. Mind the sign of $a$ ⁠.

Worked solution
1. The givens: $x_0 = 0$ ⁠, v₀ = 0.4 m/s, a = −0.1 m/s², t = 3 s.
2. With constant acceleration: $x = x_0 + v_0 t + \tfrac12 at^2$ ⁠.
3. Substitute: $x = 0 + 0.4 \times 3 + \tfrac12 \times (-0.1) \times 3^2 = 1.2 - 0.45 = 0.75$ ⁠.
4. The cart is 0.75 m past the start line.
If you got 0.3 m, you forgot the half: $1.2 - 0.9$ ⁠. If you got 1.2 m, you used the starting velocity for the whole time, as if there were no acceleration.
Step 2 of 6
Open this step
When it stops
Practice · Calculate
Question
The cart crosses the start line at 0.75 m/s, and the fan pushes back with an acceleration of −0.25 m/s². After how long does it stop?

Correct answer
3 s ± 0.05 s
Setup, choices and feedback
The Scene at this step
Starting velocity
0.75 m/sfixed
Acceleration
-0.25 m/s²fixed
Build the graphs
All three graphsfixed
Other settings
Readings/nodes/cart/props/readings = false
Accepted within
± 0.05 s
Computed from the Scene at this step’s setting
When the cart stops and turns round (NaN if it does not within the run)/nodes/cart/outputs/turnTime
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
$0 = 0.75 - 0.25t$ ⁠, so $t = 3$ ⁠ seconds: the velocity drops by $0.25$ ⁠ each second, and three seconds take away all $0.75$ ⁠.

Hint
At the moment it stops $v = 0$ ⁠. Put that into $v = v_0 + at$ ⁠ and solve for $t$ ⁠.

Worked solution
1. The givens: v₀ = 0.75 m/s, a = −0.25 m/s², and at the moment it stops $v = 0$ ⁠.
2. $v = v_0 + at$ ⁠, so $0 = 0.75 - 0.25t$ ⁠.
3. $0.25t = 0.75$ ⁠, so $t = \dfrac{0.75}{0.25} = 3$ ⁠.
4. The cart stops after 3 s, at the turning point of the position graph.
If you got −3 s, you worked out $\dfrac{v_0}{a}$ ⁠ without the minus in the equation: $t = -\dfrac{v_0}{a}$ ⁠. The stopping time is always positive, because it comes after the start.
Step 3 of 6
Open this step
Stopping on a mark
Practice · Reach the target
Question
Set up a run in which the cart stops after exactly 1.5 s, 0.45 m from the start line. Change Starting velocity and Acceleration, and press Check.

Goal
- 1.5 s ± 0.05 s
  At the start of the step: 2 s
- 0.45 m ± 0.03 m
  At the start of the step: 0.5625 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/s
Acceleration
-0.3 m/s²
Build the graphs
All three graphs
Other settings
Probe time/nodes/cart/props/probeTime = 1.5
Read from the Scene at
When the cart stops and turns round (NaN if it does not within the run)/nodes/cart/outputs/turnTime
Read from the Scene at
Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
One setting that reaches it
Starting velocity
0.6 m/s
Acceleration
-0.4 m/s²
Feedback for a right answer
The cart stops on the mark. Notice that the velocity equation fixed only the ratio of $v_0$ ⁠ to $a$ ⁠; the position fixed both.

Feedback for a wrong answer
Not yet. Check the time first: $t = -\dfrac{v_0}{a}$ ⁠ has to come out as $1.5$ ⁠. Then adjust both together until the distance fits as well.

Hint
Up to the stop the velocity falls from $v_0$ ⁠ to zero, so the average velocity is $\tfrac12 v_0$ ⁠. What must that be to cover the distance in that time?

Worked solution
1. The givens: the stop comes after t = 1.5 s, at x = 0.45 m.
2. Up to the stop the velocity falls in a straight line from $v_0$ ⁠ to zero, and the area under it is a triangle: $x = \tfrac12 v_0 t$ ⁠.
3. Substitute: $0.45 = \tfrac12 v_0 \times 1.5$ ⁠, so $v_0 = 0.6$ ⁠.
4. From $0 = v_0 + at$ ⁠: $a = -\dfrac{0.6}{1.5} = -0.4$ ⁠.
5. Set Starting velocity to 0.6 m/s and Acceleration to −0.4 m/s².
The starting run, 0.6 m/s with −0.3 m/s², stops too late, after 2 s. And 0.45 m/s with −0.3 m/s² stops at the right time, but only 0.34 m from the line.
Step 4 of 6
Open this step
Without the time
Practice · Calculate
Question
A car travelling at 20 m/s (about 72 km/h) brakes with a constant acceleration of −5 m/s². How many metres does it cover before it stops?

Correct answer
40 m ± 0.5 m
Setup, choices and feedback
The Scene at this step
Starting velocity
0.6 m/sfixed
Acceleration
-0.3 m/s²fixed
Build the graphs
All three graphs
Accepted within
± 0.5 m
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 80 m ± 0.5 m
  Misconception: Drops the ½ (or the 2) in a constant-acceleration equation: x = v₀t + at², or a stopping distance of v₀²/a.
  Feedback from the catalogue
  The half comes from the triangle under the velocity graph: base $t$ ⁠, height $at$ ⁠, and area half their product, $\tfrac12 at^2$ ⁠. In the same way $v^2 = v_0^2 + 2a\Delta x$ ⁠ gives a stopping distance of $\dfrac{v_0^2}{2|a|}$ ⁠, not $\dfrac{v_0^2}{|a|}$ ⁠. Without the half the answer comes out twice as big.
Feedback for a right answer
$0 = 20^2 + 2 \times (-5) \times \Delta x$ ⁠, so $\Delta x = \dfrac{400}{10} = 40$ ⁠ metres. Check: it stops after $4$ ⁠ seconds, at an average velocity of $10$ ⁠ metres per second.

Hint
The time is not given and not needed. Use $v^2 = v_0^2 + 2a\,\Delta x$ ⁠ with $v = 0$ ⁠.

Worked solution
1. The givens: v₀ = 20 m/s, a = −5 m/s², and at the end $v = 0$ ⁠. The time is not given.
2. An equation without time: $v^2 = v_0^2 + 2a\,\Delta x$ ⁠.
3. Substitute: $0 = 400 + 2 \times (-5) \times \Delta x = 400 - 10\,\Delta x$ ⁠.
4. $\Delta x = 40$ ⁠: a braking distance of 40 m.
If you got 80 m, you worked out $\dfrac{v_0^2}{|a|}$ ⁠ and forgot the $2$ ⁠. Another way to check: the braking lasts $\dfrac{20}{5} = 4$ ⁠ seconds, and the average velocity during it is half of $20$ ⁠, so $10 \times 4 = 40$ ⁠.
Step 5 of 6
Open this step
Twice the speed
Practice · Predict, then look
Question
At 0.3 m/s and with a backward push of −0.3 m/s², the cart stops 0.15 m from the start line. Now it will set off at 0.6 m/s, twice as fast, with the same push. How far from the start line will it stop?

Correct answer
- Correct 0.6 m, four times as far
Setup, choices and feedback
The Scene at this step
Starting velocity
0.3 m/sfixed
Acceleration
-0.3 m/s²fixed
Build the graphs
Position graph
Other settings
Probe time/nodes/cart/props/probeTime = 2
Choices
- Correct 0.6 m, four times as far
- Wrong 0.3 m, twice as far
  Misconception: Thinks the stopping distance doubles when the speed doubles.
  Feedback from the catalogue
  The stopping distance is $\dfrac{v_0^2}{2|a|}$ ⁠, so it grows with the square of the speed. Twice the speed means twice the time to stop, and over all that time an average speed twice as high too, so four times the distance.
- Wrong 0.15 m, the same place, because the push is the same
  Feedback
  The same push takes the velocity down at the same rate, but now there is more velocity to take away. That takes longer, and in that time the cart is moving faster.
Once the prediction is locked in, the Scene changes
Starting velocity
0.6 m/sfixed
The Scene then shows: 0.6 m · Position from the start mark at probeTime (held at an end stop)/nodes/cart/outputs/positionAt
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$\dfrac{0.6^2}{2 \times 0.3} = 0.6$ ⁠ metres, four times 0.15 m. The cart travels for twice as long, two seconds instead of one, at twice the average velocity.

Worked solution
1. The stopping distance: $\Delta x = \dfrac{v_0^2}{2|a|}$ ⁠.
2. At 0.3 m/s: $\dfrac{0.09}{0.6} = 0.15$ ⁠ metres. At 0.6 m/s: $\dfrac{0.36}{0.6} = 0.6$ ⁠ metres.
3. The speed went up by a factor of $2$ ⁠, and the distance by $2^2 = 4$ ⁠.
If you chose 0.3 m, you assumed the distance grows like the speed. But twice the speed also means twice the time to stop: twice as fast, for twice as long. That is why a car at 100 km/h needs four times the braking distance of a car at 50 km/h.
Step 6 of 6
Open this step
Summary
No practice question in this step.
The Scene at this step
Starting velocity
0.6 m/sdefault
Acceleration
-0.3 m/s²default
Build the graphs
Position graphdefault

Across the unit · 14 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Position and velocity

Position with a sign

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Without a push

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Velocity is the slope

Question

Correct answer

The Scene at this step

Hint

Worked solution

Negative velocity

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The equation of uniform motion

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Summary

Acceleration

A growing velocity

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A cart that is already moving

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

A push against the motion

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Negative acceleration, and a cart that speeds up

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

At the turning point

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Summary

Reading motion graphs

The graph is not a picture of the path

Question

Correct answer

The Scene at this step

Choices

Hint