M1 3.1 - Reading motion graphs

M1 3.1 - Reading motion graphs

Read motion from the gradient or area of a graph, with careful attention to the vertical axis. Distinguish displacement from distance and velocity from speed, including journeys that reverse direction.

Displacement-time slopes

A particle returns to its starting point after travelling 16 m16\,\mathrm m. How can a graph show zero displacement without claiming that nothing happened? We need to distinguish a signed position change from the total path length.

Choose a positive direction along the line of motion. Displacement ss from a fixed reference point is signed. Distance travelled adds every length of path, regardless of direction. On a displacement-time graph, moving up the page means displacement is increasing, not necessarily that the particle is physically moving upwards.

For example, a rise of 12 m12\,\mathrm m in 4 s4\,\mathrm s gives velocity 3 m s−13\,\mathrm{m\,s^{-1}}. This is the graph's gradient.

Displacement-time gradient

velocity=change in displacementelapsed time\text{velocity}=\frac{\text{change in displacement}}{\text{elapsed time}}

for a straight segment of a displacement-time graph. A horizontal segment means rest; a negative gradient means motion in the negative direction. On a curve, the tangent gradient gives instantaneous velocity. Not in the formula booklet: learn the graph interpretation.

[DIAGRAM: asset_name: Displacement-time slopes; asset_slug: m1-displacement-slopes; caption: A downward-sloping displacement graph represents negative velocity, not a negative distance travelled.; description: Two displacement-time panels show a straight line from zero seconds two metres to four seconds fourteen metres, and a straight line from zero seconds fourteen metres to four seconds two metres, with gradients plus three and minus three metres per second.]
A downward-sloping displacement graph represents negative velocity, not a negative distance travelled.

Worked example 1: positive slope

Question. A straight displacement-time segment joins (0,2)(0,2) to (4,14)(4,14), with time in seconds and displacement in metres. Find the velocity and distance travelled over the segment.

Plan. Calculate the gradient; the constant sign means there is no reversal.

v=14−24−0=3 m s−1,d=14−2=12 m.v=\frac{14-2}{4-0}=3\,\mathrm{m\,s^{-1}}, \qquad d=14-2=12\,\mathrm m.

Answer. Velocity 3 m s−13\,\mathrm{m\,s^{-1}}, distance 12 m12\,\mathrm m.

Check. Four seconds at 3 m s−13\,\mathrm{m\,s^{-1}} changes position by 12 m12\,\mathrm m, from 22 to 1414.

Reverse only the order of the displacement values. The path length stays positive.

Worked example 2: negative slope

Question. A straight displacement-time segment joins (0,14)(0,14) to (4,2)(4,2). Find the velocity and distance travelled, using the same units.

Plan. Keep final minus initial in the gradient, then take the size of the position change for distance.

v=2−144−0=−3 m s−1,d=∣2−14∣=12 m.v=\frac{2-14}{4-0}=-3\,\mathrm{m\,s^{-1}}, \qquad d=\lvert2-14\rvert=12\,\mathrm m.

Answer. Velocity −3 m s−1-3\,\mathrm{m\,s^{-1}}, distance 12 m12\,\mathrm m.

Check. 14+4(−3)=214+4(-3)=2, recovering the final displacement.

Compare heights for position and slopes for velocity.

Watch out

A higher point on a displacement-time graph does not mean a higher speed. Two parallel straight segments have the same velocity even if they are at different heights.

Use both the time difference and the signed displacement difference.

Velocity-time areas

The area between a velocity-time graph and the time axis gives displacement, with areas below the axis counted negatively.

During a short interval at constant velocity vv, displacement is vΔtv\Delta t: the area of a rectangle. For a linearly changing velocity, add thin rectangles or use the exact triangle/trapezium area. That is why the vertical-axis sign matters.

Velocity-time graphs

displacement=signed area under the velocity-time graph,\text{displacement}=\text{signed area under the velocity-time graph}, acceleration=change in velocityelapsed time\text{acceleration}=\frac{\text{change in velocity}}{\text{elapsed time}}

for a straight segment. A horizontal velocity graph means constant velocity, which may be nonzero. These graph interpretations are not printed in the formula booklet: learn them.

A straight segment from uu to vv over time tt has signed trapezium area 12(u+v)t\tfrac12(u+v)t, including when it crosses the axis. This can give zero or a negative result; neither means zero distance travelled.

Worked example 3: positive velocity throughout

Question. Velocity increases linearly from 2 m s−12\,\mathrm{m\,s^{-1}} to 8 m s−18\,\mathrm{m\,s^{-1}} over 3 s3\,\mathrm s. Find displacement and acceleration.

Plan. Use trapezium area for displacement and gradient for acceleration.

s=12(2+8)×3=15 m,a=8−23=2 m s−2.s=\frac12(2+8)\times3=15\,\mathrm m, \qquad a=\frac{8-2}{3}=2\,\mathrm{m\,s^{-2}}.

Answer. Displacement 15 m15\,\mathrm m, acceleration 2 m s−22\,\mathrm{m\,s^{-2}}.

Check. The average velocity is halfway between 22 and 88, namely 55; 5×3=155\times3=15.

Change only the final velocity to −8-8. The area now includes a negative region.

Worked example 4: crossing the time axis

Question. Velocity changes linearly from 2 m s−12\,\mathrm{m\,s^{-1}} to −8 m s−1-8\,\mathrm{m\,s^{-1}} over 3 s3\,\mathrm s. Find displacement and acceleration.

Plan. Keep the signed velocities in the trapezium and gradient calculations.

s=12(2−8)×3=−9 m,a=−8−23=−103 m s−2.s=\frac12(2-8)\times3=-9\,\mathrm m, \qquad a=\frac{-8-2}{3}=-\frac{10}{3}\,\mathrm{m\,s^{-2}}.

Answer. Displacement −9 m-9\,\mathrm m, acceleration −103 m s−2-\frac{10}{3}\,\mathrm{m\,s^{-2}}.

Check. The graph crosses zero after 0.6 s0.6\,\mathrm s. The positive triangle has area 0.6 m0.6\,\mathrm m; the negative triangle has area −9.6 m-9.6\,\mathrm m. Their sum is −9 m-9\,\mathrm m.

[DIAGRAM: asset_name: Signed area under a velocity graph; asset_slug: m1-velocity-signed-area; caption: The two triangular regions contribute with opposite signs.; description: A straight velocity-time graph falls from two metres per second at zero seconds to negative eight metres per second at three seconds. It crosses the time axis at zero point six seconds. The positive triangle is labelled plus zero point six metres and the lower triangle minus nine point six metres.]
The two triangular regions contribute with opposite signs.

The sign reports the net direction from the starting point.

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