Mechanics 7.2 - Kinematics graphs in straight-line motion

Mechanics 7.2 - Kinematics graphs in straight-line motion

Read a motion graph as information about gradients, areas and direction. You will use displacement-time and velocity-time graphs to solve straight-line journeys, including reversals and journeys with several stages.

The graph language of motion

Can a graph slope downwards while the particle still moves forwards? Yes: a velocity decreasing from 88 to 2 m s−12\,\mathrm{m\,s^{-1}} stays positive. The downward slope describes acceleration, not direction of motion.

You already know

Displacement and velocity are signed. Distance and speed are nonnegative. A straight-line gradient is rise divided by run.

Read both axis labels before giving a physical meaning to that gradient.

Kinematics graphs

A displacement-time graph plots signed displacement ss against time tt. A velocity-time graph plots signed velocity vv against tt. A speed-time graph plots ∣v∣|v|, so it cannot go below the time axis.

Units reveal what a gradient or area measures.

GraphGradientArea between graph and time axis
Displacement-timeVelocity: m/s\mathrm m/\mathrm sNo standard kinematic quantity here
Velocity-timeAcceleration: m s−1/s\mathrm{m\,s^{-1}}/\mathrm sSigned displacement: m s−1×s\mathrm{m\,s^{-1}}\times\mathrm s
Speed-timeRate of change of speedDistance travelled

A graph is a record of motion over time, not a drawing of the road. A horizontal displacement-time line means rest; a horizontal velocity-time line can mean steady travel.

Key idea

Graph height gives the vertical-axis quantity. Gradient and area give different information.

Diagnose the interpretation before calculating.

Reading displacement-time graphs

A displacement rising from 44 to 16 m16\,\mathrm m over 6 s6\,\mathrm s gives an average velocity of 2 m s−12\,\mathrm{m\,s^{-1}}. If that graph section is straight, its velocity is constant.

Velocity as a gradient

average velocity=ΔsΔt.\text{average velocity}=\frac{\Delta s}{\Delta t}.

A straight section has this constant velocity. On a curve, the tangent gradient gives instantaneous velocity and a chord gradient gives average velocity. Not in the formula booklet: learn it.

Positive, zero and negative gradients correspond to forward motion, rest and backward motion relative to the chosen positive direction.

[DIAGRAM: asset_slug: p3_7_2_understand_and_interpret_kinematics_graphs_in_straight_line_motion__diagram_01; asset_name: Displacement-time gradients; caption: Positive, zero and negative gradients give the three velocity signs; description: Three displacement-time graphs show positive gradient, a horizontal line and negative gradient, with clear axes and velocity interpretations.]
Positive, zero and negative gradients give the three velocity signs

Choose points far apart on a straight section to reduce reading error.

Worked example 1: a positive gradient

Question. A straight displacement-time section joins (2,4)(2,4) and (8,16)(8,16), with time in seconds and displacement in metres. Find velocity.

Plan. Divide the change in displacement by the change in time.

Answer.

v=16−48−2=126=2 m s−1.v=\frac{16-4}{8-2}=\frac{12}{6}=2\,\mathrm{m\,s^{-1}}.

Why subtract both coordinates? The gradient describes change over the interval, not displacement divided by the final time.

Check. In 6 s6\,\mathrm s, velocity 2 m s−12\,\mathrm{m\,s^{-1}} adds 12 m12\,\mathrm m, matching the rise.

Change only the second displacement from 1616 to −8 m-8\,\mathrm m.

Worked example 2: a negative gradient

Question. A straight section joins (2,4)(2,4) and (8,−8)(8,-8). Find velocity and interpret its sign.

Plan. Use the same coordinate differences in the same order.

Answer.

v=−8−48−2=−2 m s−1.v=\frac{-8-4}{8-2}=-2\,\mathrm{m\,s^{-1}}.

The particle moves in the negative direction. Why can the final displacement be negative? It has passed the origin; its velocity comes from slope, not height.

Check. 4+(−2)(6)=−8 m4+(-2)(6)=-8\,\mathrm m.

For a curved graph, draw or use the supplied tangent at the requested time and calculate its gradient with two points on the tangent. Those points need not lie on the curve. For example, tangent points (1,6)(1,6) and (5,14)(5,14) give instantaneous velocity (14−6)/(5−1)=2 m s−1(14-6)/(5-1)=2\,\mathrm{m\,s^{-1}}.

Watch out

Graph height is displacement. Using the height as velocity ignores the rate at which displacement changes.

Keep the first point and change only the second displacement again.

Reading velocity-time graphs

A velocity-time graph carries two pieces of information: slope tells you how velocity changes, while signed area adds the displacement accumulated during the interval.

Gradient and signed area

a=ΔvΔton a straight section,s=signed area under the velocity-time graph.a=\frac{\Delta v}{\Delta t}\quad\text{on a straight section},\qquad s=\text{signed area under the velocity-time graph}.

A tangent gives instantaneous acceleration on a curve. Areas below the axis count negatively for displacement; add absolute areas for distance. Not in the formula booklet: learn it.

The small rectangle idea explains the area rule: approximately constant velocity vv over a short interval Δt\Delta t contributes displacement vΔtv\Delta t. Adding such contributions gives area.

[DIAGRAM: asset_slug: p3_7_2_understand_and_interpret_kinematics_graphs_in_straight_line_motion__diagram_02; asset_name: Velocity-time gradient and signed area; caption: A slope measures acceleration and signed areas measure displacement; description: A gradient triangle is shown separately from a piecewise velocity-time graph with positive and negative shaded regions.]
A slope measures acceleration and signed areas measure displacement

Start with a straight graph that crosses the axis.

Worked example 3: an axis crossing

Question. Velocity decreases linearly from 6 m s−16\,\mathrm{m\,s^{-1}} at t=0t=0 to −2 m s−1-2\,\mathrm{m\,s^{-1}} at t=4t=4. Find acceleration, displacement and distance.

Plan. Find slope and the zero-velocity time, then calculate both triangle areas.

Answer.

a=−2−64=−2 m s−2gradient0=6−2t⇒t=3axis crossings=12(3)(6)−12(1)(2)=8 msigned areasD=9+1=10 mabsolute areas\begin{aligned} a&=\frac{-2-6}{4}=-2\,\mathrm{m\,s^{-2}} &&\text{gradient}\\ 0&=6-2t\quad\Rightarrow\quad t=3 &&\text{axis crossing}\\ s&=\tfrac12(3)(6)-\tfrac12(1)(2)=8\,\mathrm m &&\text{signed areas}\\ D&=9+1=10\,\mathrm m &&\text{absolute areas} \end{aligned}

Why split at t=3t=3? Velocity changes sign there, so the particle reverses.

Check. The signed trapezium area is 12(6−2)(4)=8 m\tfrac12(6-2)(4)=8\,\mathrm m.

Change only the final velocity to −6 m s−1-6\,\mathrm{m\,s^{-1}}.

Worked example 4: equal opposite areas

Question. Velocity decreases linearly from 66 to −6 m s−1-6\,\mathrm{m\,s^{-1}} over 4 s4\,\mathrm s. Find acceleration, displacement and distance.

Plan. Use symmetry for the crossing time and keep signs for displacement.

Answer.

a=−6−64=−3 m s−2trest=2 smidpoint of the straight sections=12(2)(6)−12(2)(6)=0 mD=6+6=12 m\begin{aligned} a&=\frac{-6-6}{4}=-3\,\mathrm{m\,s^{-2}}\\ t_{\mathrm{rest}}&=2\,\mathrm s &&\text{midpoint of the straight section}\\ s&=\tfrac12(2)(6)-\tfrac12(2)(6)=0\,\mathrm m\\ D&=6+6=12\,\mathrm m \end{aligned}

Why is distance nonzero? Opposite displacements cancel, but both path lengths count.

Check. The average velocity on this straight section is (6−6)/2=0(6-6)/2=0.

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