Motion graphs

AS · 12 min

A motion graph turns a description of a journey into a picture you can measure. Two rules do almost all the work: the gradient of a displacement–time graph is velocity, and the gradient of a velocity–time graph is acceleration, while the area under a velocity–time graph is displacement. Paper 1 tests these in nearly every session, and Paper 2 often opens a mechanics question with a graph to read or sketch.

Displacement–time graphs

On a displacement–time ( $s$ – $t$ ) graph, displacement is plotted on the vertical axis and time on the horizontal axis.

Key result

\text{velocity} = \text{gradient of the displacement–time graph} = \frac{\Delta s}{\Delta t}

A straight line means constant velocity. Steeper means faster.
A horizontal line means the object is at rest.
A negative gradient means the object is moving in the negative direction (back towards, or past, the reference point).
A curve means the velocity is changing. The instantaneous velocity at any moment is the gradient of the tangent to the curve at that point.

A distance–time graph looks the same except that distance can never decrease: when the object turns back, a displacement–time graph slopes down but a distance–time graph keeps rising. Its gradient is the speed.

y = 0.75 x^2 y = 6 x - 12 (4, 0) -- (4, 12)

The curve is $s = 0.75t^2$ , an object accelerating from rest. The straight line is the tangent at $t = 4\ \text{s}$ ; its gradient gives the instantaneous velocity at that moment.

Method

Gradient of a tangent to a curve

Place a ruler so it touches the curve at the point, with equal angles either side (it should not cut the curve nearby).
Draw the tangent long: extend it across most of the grid.
Pick two points on the tangent (not on the curve) that are far apart, read their coordinates, and calculate $\Delta y / \Delta x$ with units.

Instantaneous velocity from a tangent

For the graph above, the tangent at $t = 4.0\ \text{s}$ passes through $(2.0\ \text{s},\ 0\ \text{m})$ and $(6.0\ \text{s},\ 24\ \text{m})$ . Find the velocity at $t = 4.0\ \text{s}$ , and the average velocity over the first $4.0\ \text{s}$ .

Solution

Instantaneous velocity is the gradient of the tangent:

v = \frac{24 - 0}{6.0 - 2.0} = \frac{24}{4.0} = 6.0\ \text{m s}^{-1}

Average velocity uses the chord from the origin to the point on the curve, $s = 0.75 \times 4.0^2 = 12\ \text{m}$ :

\bar{v} = \frac{12}{4.0} = 3.0\ \text{m s}^{-1}

For uniform acceleration from rest, the average velocity is half the final velocity, which checks.

Velocity–time graphs

The velocity–time ( $v$ – $t$ ) graph is the most useful of the three, because it carries both acceleration and displacement.

Key result

\text{acceleration} = \text{gradient of the velocity–time graph} = \frac{\Delta v}{\Delta t}

\text{displacement} = \text{area between the velocity–time graph and the time axis}

Why is the area the displacement? Over a short time $\Delta t$ the velocity is nearly constant, so the displacement in that interval is $v\,\Delta t$ : the area of a thin strip under the graph. Adding up all the strips gives the total area. For constant velocity this is just a rectangle, $s = vt$ .

A horizontal line: constant velocity, zero acceleration.
A straight sloping line: uniform (constant) acceleration.
A curve: changing acceleration; use the gradient of a tangent for the instantaneous acceleration.
Area above the time axis is displacement in the positive direction; area below is displacement in the negative direction. Add them with signs for the total displacement; add their sizes for the total distance.

(0, 0) -- (4, 12) (4, 12) -- (10, 12) (10, 12) -- (16, 0) fill 0 4 y = 3 x

This velocity–time graph shows a car accelerating from rest to $12\ \text{m s}^{-1}$ in $4\ \text{s}$ , travelling at constant velocity for $6\ \text{s}$ , then decelerating to rest in $6\ \text{s}$ . The shaded triangle is the displacement during the first $4\ \text{s}$ .

Reading a velocity–time graph

For the car in the graph above, calculate (a) the acceleration in each stage, (b) the total displacement and (c) the average velocity for the whole journey.

Solution

(a) $0$ to $4\ \text{s}$ : $a = 12/4 = 3.0\ \text{m s}^{-2}$ . $4$ to $10\ \text{s}$ : $a = 0$ . $10$ to $16\ \text{s}$ : $a = (0 - 12)/6 = -2.0\ \text{m s}^{-2}$ .

(b) Split the area into a triangle, a rectangle and a triangle:

s = \tfrac{1}{2}(4)(12) + (6)(12) + \tfrac{1}{2}(6)(12) = 24 + 72 + 36 = 132\ \text{m}

(c)

\bar{v} = \frac{132}{16} = 8.25\ \text{m s}^{-1} \approx 8.3\ \text{m s}^{-1}

Motion with a change of direction

When velocity changes sign, the object has reversed. The classic example is a ball thrown upwards.

A ball thrown upwards

A ball is thrown vertically upwards at $15\ \text{m s}^{-1}$ and caught again at the same height. Air resistance is negligible. Take upwards as positive. (a) Sketch the velocity–time graph. (b) Use it to find the time to reach the top, the maximum height, the total distance travelled and the final displacement.

Solution

(a) The acceleration is $-9.81\ \text{m s}^{-2}$ throughout, so the graph is a single straight line of gradient $-9.81$ , from $+15\ \text{m s}^{-1}$ at $t = 0$ , through zero, to $-15\ \text{m s}^{-1}$ when caught.

y = 15 - 9.81 x fill 0 1.529 y = 15 - 9.81 x

(b) At the top, $v = 0$ : $t = 15/9.81 = 1.53\ \text{s}$ . The ball is caught at $t = 2 \times 1.53 = 3.06\ \text{s}$ .

Maximum height $=$ area of the shaded triangle above the axis:

h = \tfrac{1}{2} \times 1.53 \times 15 = 11.5\ \text{m}

The triangle below the axis has the same area, so the total distance $= 2 \times 11.5 = 22.9\ \text{m}$ , while the displacement $= 11.5 + (-11.5) = 0$ .

Watch out

At the top of its flight the ball's velocity is zero but its acceleration is still $-9.81\ \text{m s}^{-2}$ . The graph passes straight through the axis with the same gradient. If the acceleration were zero at the top, the ball would stay there.

A bouncing ball

A ball dropped onto a hard floor gives a velocity–time graph of parallel sloping lines. With upwards positive, the ball's velocity becomes more and more negative as it falls (gradient $-9.81\ \text{m s}^{-2}$ ). At impact it reverses almost instantly, jumping to a smaller positive value. It then slows (same gradient) to zero at the top of the bounce, falls again, and so on.

(0, 0) -- (0.606, -5.94) (0.606, -5.94) -- (0.606, 4.4) (0.606, 4.4) -- (1.055, 0) (1.055, 0) -- (1.503, -4.4) (1.503, -4.4) -- (1.503, 3.3) (1.503, 3.3) -- (1.84, 0)

The ball here is dropped from $1.8\ \text{m}$ , hits the floor at $5.9\ \text{m s}^{-1}$ after $0.61\ \text{s}$ , and rebounds at $4.4\ \text{m s}^{-1}$ . Every sloping line has the same gradient, $-9.81\ \text{m s}^{-2}$ . The near-vertical lines are the brief impacts, where the acceleration is very large and upwards. The area of each triangle below the axis is the height fallen; each triangle above is the height of the next bounce.

Acceleration–time graphs

The area under an acceleration–time graph is the change in velocity. For uniformly accelerated motion it is a horizontal line. Paper 1 sometimes asks you to match an acceleration–time graph to a velocity–time graph: a constant positive acceleration matches a straight rising $v$ – $t$ line; a step from positive to zero matches a $v$ – $t$ line that rises and then levels off.

Areas under curves

When the velocity–time graph is curved, find the area by:

counting squares: count whole squares under the curve, count part-squares as halves where roughly half is under, then multiply by the displacement represented by one square ( $\text{value of one square in } v \times \text{value in } t$ ); or
splitting into trapezia: divide the time axis into equal strips of width $\Delta t$ and add the trapezium areas $\tfrac{1}{2}(v_1 + v_2)\Delta t$ .

Area under a curve by trapezia

A sprinter's velocity during the first $4.0\ \text{s}$ is read from a curved graph at $1.0\ \text{s}$ intervals: $0$ , $3.15$ , $5.06$ , $6.22$ and $6.92\ \text{m s}^{-1}$ . Estimate the distance run in $4.0\ \text{s}$ , and state whether the estimate is too high or too low.

Solution

With strips of width $1.0\ \text{s}$ , the sum of trapezia is

s \approx 1.0 \times \left[\tfrac{1}{2}(0 + 6.92) + 3.15 + 5.06 + 6.22\right] = 3.46 + 14.43 = 17.9\ \text{m}

The graph curves above the straight tops of the trapezia (velocity rises quickly at first, then levels off), so each trapezium misses a sliver of area: the estimate is slightly too low. (The exact area for this graph is $18.2\ \text{m}$ .)

Sketching graphs from a description

Many questions give a description and ask for a sketch. Work stage by stage, deciding for each stage what the velocity is doing (constant, increasing, decreasing, reversing) and therefore what the gradient of each graph must be.

Motion	$s$ – $t$ graph	$v$ – $t$ graph	$a$ – $t$ graph
at rest	horizontal line	on the time axis ( $v = 0$ )	on the time axis
constant velocity	straight sloping line	horizontal line	on the time axis
uniform acceleration from rest	parabola, getting steeper	straight line through origin	horizontal line above axis
uniform deceleration to rest	curve, getting less steep, flattening	straight line down to axis	horizontal line below axis
falling with air resistance (from rest)	curve becoming a straight line	curve levelling off at terminal velocity	decreasing curve to zero

A lift journey

A lift starts from rest at the ground floor. Taking upwards as positive, it accelerates uniformly to $2.0\ \text{m s}^{-1}$ in $2.0\ \text{s}$ , travels at this speed for $5.0\ \text{s}$ , then decelerates uniformly to rest in $2.0\ \text{s}$ . (a) Sketch its velocity–time and acceleration–time graphs. (b) Find the height it rises.

Solution

(a) Velocity–time: a straight line from $(0, 0)$ to $(2.0, 2.0)$ , horizontal to $(7.0, 2.0)$ , then a straight line down to $(9.0, 0)$ .

(0, 0) -- (2, 2) (2, 2) -- (7, 2) (7, 2) -- (9, 0)

Acceleration–time: $+1.0\ \text{m s}^{-2}$ from $0$ to $2.0\ \text{s}$ , zero from $2.0$ to $7.0\ \text{s}$ , $-1.0\ \text{m s}^{-2}$ from $7.0$ to $9.0\ \text{s}$ .

(0, 1) -- (2, 1) (2, 0) -- (7, 0) (7, -1) -- (9, -1)

(b) Area under the $v$ – $t$ graph (a trapezium):

h = \tfrac{1}{2}(5.0 + 9.0) \times 2.0 = 14\ \text{m}

Exam tip

Always give the unit of a gradient or area: work it out as (unit on $y$ -axis)/(unit on $x$ -axis) or (unit on $y$ -axis) $\times$ (unit on $x$ -axis). Check the axes for prefixes such as km or ms.
When finding a gradient of a straight line, use a large triangle (more than half the line). For a tangent, draw it long and use points on the tangent.
"Determine the displacement" from a graph with parts below the axis needs the signed sum; "determine the distance" needs the sum of sizes. Read which word is used.
For sketches, the examiner looks for the right shape in each stage, correct sign, and key values marked on the axes (for example the terminal velocity, or where $v = 0$ ).

Summary

Gradient of $s$ – $t$ = velocity; gradient of $v$ – $t$ = acceleration.
Area under $v$ – $t$ = displacement (signed); area under $a$ – $t$ = change in velocity.
Instantaneous values on curves come from the gradient of a tangent.
Area below the time axis is negative displacement; add sizes for distance.
For projectiles and bounces with up positive, the $v$ – $t$ graph has gradient $-9.81\ \text{m s}^{-2}$ in free flight.
At the top of a vertical throw, $v = 0$ but $a = -9.81\ \text{m s}^{-2}$ .
Estimate curved areas by counting squares or trapezia, and say whether the estimate is high or low.

Practice

Question

State what is represented by (a) the gradient of a displacement–time graph, (b) the area under a velocity–time graph, (c) the gradient of a velocity–time graph.
A cyclist moves at $6.0\ \text{m s}^{-1}$ for $20\ \text{s}$ , stops for $10\ \text{s}$ , then returns to the start at $4.0\ \text{m s}^{-1}$ . Sketch the displacement–time graph and find the total time.
A car accelerates uniformly from $5.0\ \text{m s}^{-1}$ to $25\ \text{m s}^{-1}$ in $8.0\ \text{s}$ . Use the area under the velocity–time graph to find the distance travelled.
A velocity–time graph is a straight line from $(0\ \text{s},\ 20\ \text{m s}^{-1})$ to $(10\ \text{s},\ -10\ \text{m s}^{-1})$ . Calculate (a) the acceleration, (b) the time at which the object is momentarily at rest, (c) the displacement after $10\ \text{s}$ and (d) the distance travelled in $10\ \text{s}$ .
Describe the velocity–time graph of a ball dropped from rest that bounces several times, each bounce lower than the last, taking upwards as positive.
A train accelerates from rest at $0.50\ \text{m s}^{-2}$ for $40\ \text{s}$ , travels at constant speed for $3.0$ minutes, then decelerates uniformly to rest in $50\ \text{s}$ . Calculate the total distance travelled and the average speed.
On a displacement–time graph for an accelerating car, the tangent at $t = 5.0\ \text{s}$ passes through $(1.0\ \text{s},\ 0\ \text{m})$ and $(9.0\ \text{s},\ 96\ \text{m})$ . Find the velocity at $t = 5.0\ \text{s}$ .
Explain why the area under a velocity–time graph represents displacement.
A sprinter accelerates uniformly from rest to $10\ \text{m s}^{-1}$ in $4.0\ \text{s}$ and then runs at this speed. How long does it take to run $100\ \text{m}$ ?
Two cars start side by side. Car X moves at a constant $20\ \text{m s}^{-1}$ . Car Y starts from rest at the same instant with a constant acceleration of $2.5\ \text{m s}^{-2}$ . Using a velocity–time sketch, find (a) the time at which the cars have the same velocity and the distance between them then, and (b) the time at which Y overtakes X.

Answers

(a) Velocity. (b) Displacement. (c) Acceleration.
Rising straight line to $120\ \text{m}$ at $20\ \text{s}$ ; horizontal to $30\ \text{s}$ ; falling straight line back to $0$ , taking $120/4.0 = 30\ \text{s}$ , reaching zero at $60\ \text{s}$ . Total time $60\ \text{s}$ .
Trapezium: $s = \tfrac{1}{2}(5.0 + 25) \times 8.0 = 120\ \text{m}$ .
(a) $a = (-10 - 20)/10 = -3.0\ \text{m s}^{-2}$ . (b) $v = 0$ when $20 - 3.0t = 0$ , $t = 6.67\ \text{s}$ . (c) Area above $= \tfrac{1}{2}(6.67)(20) = 66.7\ \text{m}$ ; area below $= \tfrac{1}{2}(3.33)(10) = 16.7\ \text{m}$ ; displacement $= 66.7 - 16.7 = 50\ \text{m}$ . (d) Distance $= 66.7 + 16.7 = 83\ \text{m}$ .
Starts at zero, straight line downwards with gradient $-9.81\ \text{m s}^{-2}$ to a negative velocity; then an almost vertical jump to a smaller positive velocity at the bounce; straight line of the same gradient down through zero (top of bounce) to a negative velocity equal in size to the rebound velocity; then another jump to a still smaller positive value, and so on. All sloping lines are parallel; each successive peak is lower.
Top speed $= 0.50 \times 40 = 20\ \text{m s}^{-1}$ . Distances: $\tfrac{1}{2}(40)(20) = 400\ \text{m}$ ; $20 \times 180 = 3600\ \text{m}$ ; $\tfrac{1}{2}(50)(20) = 500\ \text{m}$ . Total $4500\ \text{m}$ in $270\ \text{s}$ ; average speed $16.7\ \text{m s}^{-1}$ .
$v = 96/8.0 = 12\ \text{m s}^{-1}$ .
In a very short time interval $\Delta t$ , the velocity is almost constant, so the displacement is $v\,\Delta t$ , which is the area of a thin strip under the graph. The total displacement is the sum of all such strips, which is the total area under the graph.
In the first $4.0\ \text{s}$ : $s = \tfrac{1}{2}(4.0)(10) = 20\ \text{m}$ . Remaining $80\ \text{m}$ at $10\ \text{m s}^{-1}$ takes $8.0\ \text{s}$ . Total $12.0\ \text{s}$ .
(a) Y reaches $20\ \text{m s}^{-1}$ at $t = 20/2.5 = 8.0\ \text{s}$ . X has gone $160\ \text{m}$ ; Y has gone $\tfrac{1}{2}(8.0)(20) = 80\ \text{m}$ . Separation $80\ \text{m}$ (the largest gap). (b) Equal displacements: $20t = \tfrac{1}{2}(2.5)t^2$ , so $t = 16\ \text{s}$ (each has travelled $320\ \text{m}$ ). On the sketch, this is where the areas under the two lines become equal.

Displacement–time graphs

Velocity–time graphs

Motion with a change of direction

A bouncing ball

Acceleration–time graphs

Areas under curves

Sketching graphs from a description

Summary

Practice

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