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Equations and Graphs of Motion

Before you start, make sure you know: Interpreting x-t and v-t graphs (Secondary Year 2)

You leave your house walking at a steady pace, speed up when you realize you're late for the Metro, and slow to a stop at the station entrance. That story can be drawn without a single word: with a graph. And from a position, velocity, or acceleration graph you can read what the object does in each segment, without knowing the equation beforehand.

x-t Graph: The Slope Is the Velocity

In a 100 m race, one runner covers $10\ \text{m}$ every $2\ \text{s}$ and another covers $18\ \text{m}$ in the same interval. How can you tell at a glance who is moving faster?

In the $x(t)$ graph, notice how steep the line is; in the $v(t)$ graph, check that it is a horizontal line, because the velocity does not change.

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The $x$-$t$ graph of a moving object is a horizontal line at $x = 5\ \text{m}$ for $10\ \text{s}$. It is shown in the figure. a) What is the state of the object during…

The $x$-$t$ graph of a moving object is a horizontal line at $x = 5\ \text{m}$ for $10\ \text{s}$. It is shown in the figure.

t (s) x (m) 5 10 x = 5 m
  1. a) What is the state of the object during this time?
  2. b) What is its velocity? Justify your answer using the gradient of the graph.
Solve
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The $x$-$t$ graph of a pedestrian is a straight line passing through the points $(0, 2)$ and $(4, 10)$, with $x$ in meters and $t$ in seconds. It is shown in the figure.…

The $x$-$t$ graph of a pedestrian is a straight line passing through the points $(0, 2)$ and $(4, 10)$, with $x$ in meters and $t$ in seconds. It is shown in the figure.

t (s) x (m) (0, 2) (4, 10) 10 2 4 slope = ?
  1. a) Calculate the slope of the line and state what physical quantity it represents.
  2. b) Write the position equation for the pedestrian.
Solve
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A classmate shows you two $x$-$t$ graphs: in the first, the position increases along a sloped straight line; in the second, the position increases along a curve that…

A classmate shows you two $x$-$t$ graphs: in the first, the position increases along a sloped straight line; in the second, the position increases along a curve that gets steeper and steeper. They are shown in the figure.

x (m) t (s) x (m) t (s)
  1. a) Which one corresponds to uniform motion and which to accelerated motion?
  2. b) Explain what happens to the velocity in each case.
Solve

v-t Graph: The Slope Is the Acceleration

Two cars read $90\ \text{km/h}$: one is holding a fixed speed on the highway and the other has just stepped on the gas to pass. How can a graph show the difference between keeping your speed and gaining speed every second?

Since $a \neq 0$, the $v(t)$ graph is no longer horizontal: it is a sloped line, and the $x(t)$ graph curves.

Velocity-time graph: a horizontal line represents uniform motion (a = 0) and a sloped line rising from the origin represents uniformly accelerated motion (a greater than 0) v t 0 Uniform motion: a = 0 Accelerated motion: a > 0 Δt Δv slope = a = Δv / Δt
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The $v$-$t$ graph of a cyclist is a horizontal line at $v = 6\ \text{m/s}$ from $t = 0$ to $t = 5\ \text{s}$. It is shown in the figure. a) What type of motion does it…

The $v$-$t$ graph of a cyclist is a horizontal line at $v = 6\ \text{m/s}$ from $t = 0$ to $t = 5\ \text{s}$. It is shown in the figure.

t (s) v (m/s) 6 5 v = 6 m/s area = distance = ?
  1. a) What type of motion does it describe?
  2. b) Calculate the distance travelled using the area under the graph.
Solve
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The $v$-$t$ graph of a car is a straight line joining the points $(0, 0)$ and $(4, 12)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure. a)…

The $v$-$t$ graph of a car is a straight line joining the points $(0, 0)$ and $(4, 12)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

t (s) v (m/s) (4, 12) 12 4 gradient = a = ?
  1. a) Calculate the gradient of the line and state what physical quantity it represents.
  2. b) What is its value along with its unit?
Solve
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The $v$-$t$ graph of a car is a straight line joining the points $(0, 20)$ and $(5, 0)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure. a) Is…

The $v$-$t$ graph of a car is a straight line joining the points $(0, 20)$ and $(5, 0)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

t (s) v (m/s) (0, 20) 20 5 gradient = a = ?
  1. a) Is the car accelerating or braking? Justify your answer.
  2. b) Calculate its acceleration from the gradient.
Solve

Area Under v-t: Displacement and Distance

If $v<0$, the area contributes a negative displacement. For the distance traveled, add the absolute values of the areas, separating the segments where the sign of $v$ changes. For example, $v=4\ \text{m/s}$ for $3\ \text{s}$ and $v=-2\ \text{m/s}$ for $2\ \text{s}$ give $\Delta x=12-4=8\ \text{m}$, but $d=12+4=16\ \text{m}$. In the following examples the velocities are positive, so distance and displacement are the same.

Velocity-time graph with the shaded area under the line representing the object's displacement tv d = v · t (rectangle)
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The $v$-$t$ graph of a car has three sections: from $(0, 0)$ to $(10, 20)$, from $(10, 20)$ to $(30, 20)$, and from $(30, 20)$ to $(40, 0)$, with $v$ in $\text{m/s}$ and…

The $v$-$t$ graph of a car has three sections: from $(0, 0)$ to $(10, 20)$, from $(10, 20)$ to $(30, 20)$, and from $(30, 20)$ to $(40, 0)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

t (s) v (m/s) (10, 20) (30, 20) 20 10 30 40 d₁ = ? d₂ = ? d₃ = ?
  1. a) Calculate the acceleration for each section.
  2. b) Calculate the distance travelled in each section using the areas (triangles and rectangle) under the graph.
  3. c) What is the total distance?
Solve
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The $v$-$t$ graph of a car is a straight line joining the points $(0, 5)$ and $(6, 17)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure. a)…

The $v$-$t$ graph of a car is a straight line joining the points $(0, 5)$ and $(6, 17)$, with $v$ in $\text{m/s}$ and $t$ in seconds. It is shown in the figure.

t (s) v (m/s) (0, 5) (6, 17) 5 17 6 rectangle triangle
  1. a) Calculate the acceleration.
  2. b) Write the velocity equation and calculate $v$ when $t = 3\ \text{s}$.
  3. c) Calculate the distance travelled in $6\ \text{s}$ by dividing the area under the graph into a rectangle and a triangle.
Solve
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The $v$-$t$ graph of a cyclist has three straight sections connecting the points $(0, 0)$, $(5, 10)$, $(15, 10)$ and $(20, 0)$, with $v$ in $\text{m/s}$ and $t$ in…

The $v$-$t$ graph of a cyclist has three straight sections connecting the points $(0, 0)$, $(5, 10)$, $(15, 10)$ and $(20, 0)$, with $v$ in $\text{m/s}$ and $t$ in seconds.

  1. a) Describe in words what the cyclist does in each section.
  2. b) Calculate the total distance travelled using the areas under the graph.
Solve
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The $v$-$t$ graph of a metro train between two stations is: it accelerates from $0$ to $15\ \text{m/s}$ in $30\ \text{s}$, maintains $15\ \text{m/s}$ for $60\ \text{s}$…

The $v$-$t$ graph of a metro train between two stations is: it accelerates from $0$ to $15\ \text{m/s}$ in $30\ \text{s}$, maintains $15\ \text{m/s}$ for $60\ \text{s}$ and brakes to a stop at $15\ \text{s}$. It is shown in the figure.

t (s) v (m/s) 15 30 90 105 d₁ = ? d₂ = ? d₃ = ?
  1. a) Calculate the acceleration for each section.
  2. b) Calculate the distance of each section and the total distance between the stations.
  3. c) Calculate the average speed for the journey, rounded to one decimal place.
Solve

a-t Graph: The Area Is the Change in Velocity

A Metro train starts with a fixed acceleration of $0.8\ \text{m/s}^2$ for $20\ \text{s}$ and then the acceleration drops to zero. How do we find its final velocity from the graph of its acceleration?

Acceleration-time graph highlighting the rectangular area as the change in velocity ta Δv = a · t (rectangle)
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The $a$-$t$ graph of a moving object is a horizontal line at $a = 2\ \text{m/s}^2$ from $t = 0$ to $t = 6\ \text{s}$, as shown in the figure. If the object starts from…

The $a$-$t$ graph of a moving object is a horizontal line at $a = 2\ \text{m/s}^2$ from $t = 0$ to $t = 6\ \text{s}$, as shown in the figure.

t (s) a (m/s²) 2 6 a = 2 m/s² area = Δv = ?

If the object starts from rest, calculate its velocity after $6\ \text{s}$ using the area under the $a$-$t$ graph.

Solve
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The $a$-$t$ graph of an object starting from rest is: $a = 3\ \text{m/s}^2$ from $t = 0$ to $t = 4\ \text{s}$, and $a = 0$ from $t = 4\ \text{s}$ to $t = 10\ \text{s}$.…

The $a$-$t$ graph of an object starting from rest is: $a = 3\ \text{m/s}^2$ from $t = 0$ to $t = 4\ \text{s}$, and $a = 0$ from $t = 4\ \text{s}$ to $t = 10\ \text{s}$. It is shown in the figure.

t (s) a (m/s²) 3 4 10 area = Δv = ?
  1. a) Calculate the velocity at $4\ \text{s}$ using the area under the $a$-$t$ graph.
  2. b) What is its velocity at $10\ \text{s}$? Justify your answer.
  3. c) Draw the corresponding $v$-$t$ graph in your notebook.

You can do it on paper and send a photo with your answer.

Solve
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An object starts from rest. Its $a$-$t$ graph is: $a = 4\ \text{m/s}^2$ from $t = 0$ to $t = 5\ \text{s}$, and $a = -2\ \text{m/s}^2$ from $t = 5\ \text{s}$ to…

An object starts from rest. Its $a$-$t$ graph is: $a = 4\ \text{m/s}^2$ from $t = 0$ to $t = 5\ \text{s}$, and $a = -2\ \text{m/s}^2$ from $t = 5\ \text{s}$ to $t = 15\ \text{s}$. It is shown in the figure.

t (s) a (m/s²) 4 −2 0 5 15 area₁ = ? area₂ = ?
  1. a) Calculate the maximum velocity it reaches and at what time.
  2. b) Calculate the velocity at $15\ \text{s}$.
  3. c) Draw the $v$-$t$ graph in your notebook and calculate the total distance travelled using its area.
  4. d) Describe the complete motion in words.

You can do it on paper and send a photo with your answer.

Solve

Combined Segments: Rest, Uniform Motion, and Uniform Acceleration

Position-time graph with combined segments representing moving forward at constant velocity, rest, and return to the starting point constant velocity stopped returns

Two Objects on the Same Graph: The Intersection Point

Practice★★☆Not started
The $x$-$t$ graph of a walking trip has three straight sections: from $(0, 0)$ to $(10, 50)$, from $(10, 50)$ to $(15, 50)$, and from $(15, 50)$ to $(25, 0)$, with $x$…

The $x$-$t$ graph of a walking trip has three straight sections: from $(0, 0)$ to $(10, 50)$, from $(10, 50)$ to $(15, 50)$, and from $(15, 50)$ to $(25, 0)$, with $x$ in meters and $t$ in seconds. It is shown in the figure.

t (s) x (m) (10, 50) (15, 50) 50 10 15 25 v₁ = ? v₂ = ? v₃ = ?
  1. a) Calculate the speed for each section.
  2. b) Interpret each section: what are the walkers doing in each one?
  3. c) How long does the entire trip last?
Solve
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In a joint $x$-$t$ graph, object A follows the straight line $x_A = 4 \times t$ and object B follows the straight line $x_B = 20 + t$, in International System units. a)…

In a joint $x$-$t$ graph, object A follows the straight line $x_A = 4 \times t$ and object B follows the straight line $x_B = 20 + t$, in International System units.

  1. a) Draw the two straight lines in your notebook.
  2. b) What does the point where they intersect mean?
  3. c) Calculate the time and position of intersection, rounded to one decimal place.

You can do it on paper and send a photo with your answer.

Solve
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On the same $v$-$t$ graph, car A is a horizontal line at $v = 20\ \text{m/s}$ and car B is a straight line starting from the origin with acceleration…

On the same $v$-$t$ graph, car A is a horizontal line at $v = 20\ \text{m/s}$ and car B is a straight line starting from the origin with acceleration $a = 4\ \text{m/s}^2$, as shown in the figure. Both start from the same point at the same time.

t (s) v (m/s) A B t = ? 20
  1. a) Calculate the time at which the two lines on the graph cross.
  2. b) Calculate the distance travelled by each car up to that time, using the areas. Are they together?
  3. c) Calculate the time at which B actually catches up with A by equating the position equations.
  4. d) Explain the difference between "crossing of $v$-$t$ graphs" and "catching up": what is equal in each case?
Solve
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The $v$-$t$ graph of an object has two sections: a straight line from $(0, 0)$ to $(4, 8)$ and a horizontal line at $v = 8\ \text{m/s}$ from $t = 4\ \text{s}$ to…

The $v$-$t$ graph of an object has two sections: a straight line from $(0, 0)$ to $(4, 8)$ and a horizontal line at $v = 8\ \text{m/s}$ from $t = 4\ \text{s}$ to $t = 10\ \text{s}$, with $v$ in $\text{m/s}$.

  1. a) Calculate the distance travelled in each section.
  2. b) Calculate the position at $t = 2\ \text{s}$, $t = 4\ \text{s}$ and $t = 10\ \text{s}$ (take the origin as the initial position).
  3. c) Using these values, describe the shape of the $x$-$t$ graph: which section is curved and which section is straight? Justify your answer.
Solve
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The $x$-$t$ graph of an object is a straight line connecting the points $(0, 50)$ and $(10, 0)$, with $x$ in meters and $t$ in seconds. It is shown in the figure. a)…

The $x$-$t$ graph of an object is a straight line connecting the points $(0, 50)$ and $(10, 0)$, with $x$ in meters and $t$ in seconds. It is shown in the figure.

t (s) x (m) (0, 50) (10, 0) 50 −50 0 10 slope = v = ?
  1. a) Write its position equation.
  2. b) Explain what the initial position and the sign of the slope mean.
  3. c) Calculate the position when $t = 14\ \text{s}$ and explain the physical meaning of it being negative.
  4. d) At what time is it at $x = -50\ \text{m}$?
Solve
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The $x$-$t$ graph of a walk is: a straight line from $(0, 0)$ to $(4, 40)$ and another from $(4, 40)$ to $(8, 0)$, with $x$ in meters and $t$ in seconds. It is shown in…

The $x$-$t$ graph of a walk is: a straight line from $(0, 0)$ to $(4, 40)$ and another from $(4, 40)$ to $(8, 0)$, with $x$ in meters and $t$ in seconds. It is shown in the figure.

t (s) x (m) (4, 40) 40 4 8 v₁ = ? v₂ = ?
  1. a) Calculate the velocity for each section.
  2. b) Calculate the total distance traveled and the total displacement.
  3. c) Calculate the average velocity (displacement divided by time) and the average speed (distance divided by time).
  4. d) Explain why one of them comes out as zero and the other does not.
Solve
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A classmate draws a $x$-$t$ graph in which one section is a vertical line and claims that it represents an instantaneous change in position. a) Explain why a $x$-$t$…

A classmate draws a $x$-$t$ graph in which one section is a vertical line and claims that it represents an instantaneous change in position.

  1. a) Explain why a $x$-$t$ graph cannot have a vertical section: what would this imply about the position of the object at a single instant and about its velocity?
  2. b) For the same reason, explain whether the $x$-$t$ graph can "go backward" to the left on the time axis.
  3. c) Draw a correct $x$-$t$ graph in your notebook for this situation: a person leaves home, stops for a while to talk to a neighbor, and returns home. Justify the shape of each section.
Solve

Working Out the Equation from Data or a Graph

Position-time graph with experimental points lined up along a straight line of uniform motion tx aligned points = uniform motion
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Measuring the position of a laboratory trolley gives these data: at $t = 0\ \text{s}$, $x = 3\ \text{m}$; at $t = 1\ \text{s}$, $x = 5\ \text{m}$; at $t = 2\ \text{s}$,…

Measuring the position of a laboratory trolley gives these data: at $t = 0\ \text{s}$, $x = 3\ \text{m}$; at $t = 1\ \text{s}$, $x = 5\ \text{m}$; at $t = 2\ \text{s}$, $x = 7\ \text{m}$; at $t = 3\ \text{s}$, $x = 9\ \text{m}$.

  1. a) Plot the points on a $x$-$t$ graph in your notebook and check that they lie on a straight line.
  2. b) Deduce the position equation for the trolley.
  3. c) Predict its position when $t = 10\ \text{s}$.

You can do it on paper and send a photo with your answer.

Solve
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The position of a moving object, in SI units, is $x = 10 - 3 \times t$. a) Calculate its position at $t = 0\ \text{s}$, $t = 1\ \text{s}$ and $t = 2\ \text{s}$. b)…

The position of a moving object, in SI units, is $x = 10 - 3 \times t$.

  1. a) Calculate its position at $t = 0\ \text{s}$, $t = 1\ \text{s}$ and $t = 2\ \text{s}$.
  2. b) Describe the $x$-$t$ graph: is it sloping upwards or downwards? What does the negative gradient indicate?
  3. c) Calculate the time at which it passes through the origin, rounding to one decimal place.
Solve
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Measuring the position of a trolley going down an inclined plane gives: at $t = 0\ \text{s}$, $x = 0\ \text{m}$; at $t = 1\ \text{s}$, $x = 1\ \text{m}$; at…

Measuring the position of a trolley going down an inclined plane gives: at $t = 0\ \text{s}$, $x = 0\ \text{m}$; at $t = 1\ \text{s}$, $x = 1\ \text{m}$; at $t = 2\ \text{s}$, $x = 4\ \text{m}$; at $t = 3\ \text{s}$, $x = 9\ \text{m}$.

  1. a) Verify that the data satisfy the relation $x = t^2$ and, by comparing it with $x = \frac{1}{2} a \times t^2$, deduce the acceleration of the trolley.
  2. b) Predict its position when $t = 5\ \text{s}$.
  3. c) Calculate how far it moves in each second ($0$-$1\ \text{s}$, $1$-$2\ \text{s}$, $2$-$3\ \text{s}$) and explain, from these values, why the $x$-$t$ graph is a curve and not a straight line.
Solve

Investigating with Data

State the hypothesis "the cart keeps a constant velocity." Record a slow movement on a stable table, with a visible scale and the phone camera fixed, perpendicular to the path. Divide up the tasks of measuring, recording, and checking. You can also export positions from a simulation, labeling them as simulated data.

In a spreadsheet, put $t$ in column A and $x$ in column B; in C3 type =(B3-B2)/(A3-A2) and copy it down. Plot the points, fit a line, and check the differences between what was measured and what was predicted. State the units, the resolution, and possible systematic errors.

Back to the beginning: on a $v$-$t$ graph, walking at a steady pace is drawn as a horizontal line, speeding up as a rising line, and slowing down as a falling line until it touches the time axis ($v=0$).

Summary

See all the exercises in this section