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5th Year OL · Physics
Interaction · Kinematics

Equations and graphs of motion

A story: "you leave home for school at a steady pace, speed up when you realise you are late for the bus, and slow down at the school gate". That story can be drawn without a single word, using a graph. And from a graph — of position, velocity or acceleration — you can read what type of motion there is in each stage, without knowing the equation beforehand.

x-t graph: the gradient is the velocity

At a school sports day in Athlone, a stopwatch times two runners: one covers $10\ \mathrm{m}$ every $2\ \mathrm{s}$ and the other $18\ \mathrm{m}$ in the same interval. How can we compare at a glance, on a pair of coordinate axes, who is moving faster?

On the $x(t)$ graph, look at how steep the line is; on 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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Complete the following sentences about motion graphs: a) The gradient of a $x$-$t$ graph represents the $\dots$ b) The gradient of a $v$-$t$ graph represents the $\dots$…

Complete the following sentences about motion graphs:

  1. a) The gradient of a $x$-$t$ graph represents the $\dots$
  2. b) The gradient of a $v$-$t$ graph represents the $\dots$
  3. c) The area under a $v$-$t$ graph represents the distance $\dots$ travelled.
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 metres 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 metres and $t$ in seconds. It is shown in the figure.

t (s) x (m) (0, 2) (4, 10) 10 2 4 gradient = ?
  1. a) Calculate the gradient 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 gradient is the acceleration

Two cars on the M50 show $100\ \mathrm{km\,h^{-1}}$ on their dashboards: one is travelling in a straight line at a fixed velocity and the other has just put its foot down to overtake. How do we show in a graph the difference between maintaining velocity and gaining it second by second?

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

Velocity-time graph: a horizontal line represents constant velocity (a = 0) and a sloping line rising from the origin represents constant acceleration (a > 0) v t 0 Constant velocity: a = 0 Constant acceleration: a > 0 Δt Δv gradient = a = Δv / Δt
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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
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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

Signed area: displacement; absolute areas: distance travelled

If $v<0$, the area contributes a negative displacement. For the distance travelled, add the absolute values of the areas, separating the stages in which the sign of $v$ changes. For example, $v=4\ \mathrm{m\,s^{-1}}$ for $3\ \mathrm{s}$ and $v=-2\ \mathrm{m\,s^{-1}}$ for $2\ \mathrm{s}$ give $\Delta x=12-4=8\ \text{m}$, but $d=12+4=16\ \text{m}$. In the following examples the velocities are positive, which is why distance travelled and displacement coincide.

Velocity-time graph with the shaded area under the line representing the displacement of the object tv d = v · t (rectangle)
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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 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 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
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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

a-t graph: the area is the change in velocity

An Irish Rail train sets off with a constant acceleration of $0.8\ \mathrm{m\,s^{-2}}$ for $20\ \mathrm{s}$ and then the acceleration drops to zero. How do we work out the final velocity reached from the drawing of its acceleration against time?

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 stages: rest, constant velocity and constant acceleration on one graph

Displacement-time graph with combined stages representing moving forward at constant velocity, rest and returning to the starting point constant velocity stationary returns
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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 metres 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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The $x$-$t$ graph of an object is a straight line connecting the points $(0, 50)$ and $(10, 0)$, with $x$ in metres 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 metres and $t$ in seconds. It is shown in the figure.

t (s) x (m) (0, 50) (10, 0) 50 −50 0 10 gradient = v = ?
  1. a) Write its position equation.
  2. b) Explain what the initial position and the sign of the gradient 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 metres 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 metres 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 travelled 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 backwards" 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 neighbour, and returns home. Justify the shape of each section.

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

Solve

Working out the equation from data or from a graph

Two straight-line graphs meet where their equations give the same $x$ at the same $t$. For example, if $x_A = 3t$ and $x_B = 12 + t$, set $3t = 12 + t$, so $t = 6\ \mathrm{s}$ and $x = 18\ \mathrm{m}$: that point is where the two objects are at the same place at the same time.

Displacement-time graph with experimental points lined up along a straight line of constant velocity tx points in a line = constant velocity
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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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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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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 and estimating uncertainty

Put forward the hypothesis "the trolley keeps a constant velocity". Film a slow movement on a stable table, with a visible scale and a fixed camera perpendicular to the path. Share out the tasks of measuring, recording and checking; do not film faces or publish personal data. You can also export positions from a simulation, identifying them as simulated data. In the lab you could instead use a ticker-timer or light gates and a data logger.

In a spreadsheet, put $t$ in column A and $x$ in column B; in C3 type =(B3-B2)/(A3-A2) and copy it downwards. Plot the x-t points, draw a line of best fit and check the differences between the measurements and the prediction. State the units, resolution, repetitions and possible systematic errors. The relative uncertainty of $v=(0.29\pm0.02)\ \mathrm{m\,s^{-1}}$ is approximately $0.02/0.29\times100=7\%$; it is not a known true error.

Some key ideas for handling data. A random error varies unpredictably from one measurement to the next (for example, reaction time with a stopwatch); repeating and averaging reduces it. A systematic error shifts every reading the same way (for example, a wrongly set zero); repeating does not remove it. Accuracy is how close a result is to the true value; precision is how close repeated readings are to each other, so you can be precise but not accurate. Give answers to a sensible number of significant figures, no more than your least precise measurement allows: $0.29\ \mathrm{m\,s^{-1}}$ has 2 significant figures, so do not quote $0.2933\ \mathrm{m\,s^{-1}}$. A line of best fit is drawn so that the points are spread evenly above and below it; its gradient gives the velocity on an $x$-$t$ graph, and a point far from it is a possible anomalous value to investigate.

Check your interpretation against a source identified by author, date and procedure. Present the data, the conclusion and the limitations; a discordant point is investigated, not deleted to improve the fit.

Now test the model and compare it with your data:

For the free-fall report (DIFICIL-09): time 5 drops from a fixed height $h$, find the mean time $\bar{t}$ and its uncertainty (half the range), then use $h = \frac12 g t^2$, so $g = 2h/\bar{t}^2$, and compare with $9.8\ \mathrm{m\,s^{-2}}$.

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Unit challenge. Design a kinematics report integrating motions from your daily environment: 1. Free fall: Measure a fixed height ($h \approx 2.00 \pm 0.01\text{ m}$) and…

Unit challenge. Design a kinematics report integrating motions from your daily environment:

  1. Free fall: Measure a fixed height ($h \approx 2.00 \pm 0.01\text{ m}$) and time $5$ drops of a small object. Tabulate the times, find the mean and uncertainty ($\bar{t} \pm \Delta t$), experimentally calculate the value of $g$ using $h = \frac{1}{2}gt^2$, and plot the $y-t$ graph predicted by the model $y = h - \frac{1}{2}gt^2$ with your value of $g$.

  2. UFM and braking (UAM): Measure on Google Maps the distance of a journey you usually make by bike or scooter. Model an intermediate section at constant speed $v$ (UFM) and a final deceleration to a complete stop with acceleration $a$ (UAM). Write their equations of motion $x(t)$ and $v(t)$.

  3. Rotation (UCM): For the wheels ($R = 0.32\text{ m}$) spinning at that speed $v$, determine the angular velocity $\omega$, the period $T$, and the centripetal acceleration $a_c$.

Present the results in a short report with your data, calculations, and graphs.

Evaluation criteria:

  • Rigor in taking home measurements, estimating uncertainty, and plotting graphs.

  • Correctness in formulating the equations and SI units for UFM, UAM, free fall, and UCM.

  • Critical analysis of possible sources of error and clarity in the conclusions.

Solve

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

Summary

See all the exercises in this section