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ESO Y4A · Physics and Chemistry
Interaction · Kinematics

Equations and graphs of motion

A story: "you leave home at a steady pace, speed up when you realise you're late, and slow down at the end". That story can be drawn without a single word, using a graph. And from a graph you can tell, just by looking at it, what type of motion there is in each stage.

Before you start, make sure you know: Interpreting x-t and v-t graphs (3rd ESO)
Before you start, make sure you know: Linear functions (3rd ESO)

x-t graph: the gradient is the velocity

A runner passes the $10\text{ m}$ mark at $2\text{ s}$ and the $25\text{ m}$ mark at $5\text{ s}$: how can we plot these data so that we can tell at a glance whether they keep up the pace or are accelerating?

In the $x(t)$ graph, look at 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 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

v-t graph: the gradient is the acceleration

A motorbike's speedometer reads $20\text{ km/h}$, three seconds later it reads $40\text{ km/h}$ and after six seconds $60\text{ km/h}$: how do we draw these changes so we can tell whether the acceleration is constant or changes in jerks?

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 comparing a horizontal line for MRU, with zero acceleration, with a rising sloping line for MRUA, with positive acceleration. t (s) v (m/s) 0 MRUA: a > 0 sloping line MRU: a = 0 horizontal line
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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 $x$-$t$ 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

Signed area: displacement; absolute areas: distance

If $v<0$, the area contributes a negative displacement. For the distance, add up the absolute values of the areas, separating the stages in which 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 coincide.

Velocity-time graph with constant velocity: the area of the rectangle under the horizontal line, between the line and the time axis, is the displacement d = v · t. t (s) v (m/s) v t 0 Area = d d = v · t (rectangle) constant v
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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 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

Combined stages and curves

Position-time graph with three straight stages: a rising MRU line moving away from the origin, a horizontal stage at rest and a falling line of negative velocity returning to the origin. x (m) t (s) 50 0 10 15 25 MRU: moving away stationary (v = 0) returning (v < 0)
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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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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

Working out the equation from data

Position-time graph showing experimental points lying in a line, used to obtain the gradient of the line and the initial position. tx aligned points = MRU
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The position of a moving object, in SI units, is $x = 10 - 3 \cdot t$. 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 $x = 10 - 3 \cdot t$.

  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 $x = \frac{1}{2} a \cdot t^2$, 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

State 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 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, fit a straight line and check the differences between the measurements and the prediction. State the units, resolution, repetitions and possible systematic errors. The relative uncertainty of $v=(0,29\pm0,02)\ \text{m/s}$ is approximately $0,02/0,29\cdot100=7\%$; it is not a known true error.

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

Now try the model and compare it with your data:

Back to the beginning: In the $v$-$t$ graph of the story, the steady-pace stage is a horizontal line ($a=0$), the speeding up is a rising straight line ($a>0$) and the final slowing down is a straight line falling until it touches the time axis ($v=0$).

Summary

See all the exercises in this section