Mk1–Mk4 - Motion language and kinematics graphs

Mk1–Mk4 - Motion language and kinematics graphs

Distinguish motion quantities and connect position, displacement, distance, velocity, speed and acceleration graphs. Use signed quantities to describe journeys and find one object's velocity as observed from another on the same line.

Position, displacement and distance

Can a moving car have zero average velocity? Drive along a straight road and return to the starting place: the journey has positive length but zero overall displacement. The same endpoint can hide a very different journey.

You already know

A scalar has magnitude; a vector has direction as well. Along a line, choose a positive direction and an origin, then represent vectors by signed components.

Suppose a trolley starts 2 m2\,\mathrm m left of a marker, moves 9 m9\,\mathrm m right and 4 m4\,\mathrm m left. Its final coordinate is +3 m+3\,\mathrm m, its coordinate change is +5 m+5\,\mathrm m, and its route length is 13 m13\,\mathrm m.

Four different measures

Position locates the particle relative to an origin. Displacement is final position minus initial position. Distance from the origin is the magnitude of position. Distance travelled adds all route lengths positively. Position and displacement are signed on a line; distances are non-negative.

Write rr for position. The difference rfinal−rinitialr_{\text{final}}-r_{\text{initial}} remains the same if the origin is moved. Average rates must use the whole elapsed time, including stops.

Average motion measures

For elapsed time T>0T>0,

average velocity=overall displacementT,average speed=distance travelledT\text{average velocity}=\frac{\text{overall displacement}}{T},\qquad \text{average speed}=\frac{\text{distance travelled}}{T}

Not in the formula sheet: learn it.

The following calculations keep coordinate, change and travel separate.

Worked example 1: signed movement

Question. A trolley starts at −2 m-2\,\mathrm m, moves 9 m9\,\mathrm m right and 4 m4\,\mathrm m left in 5 s5\,\mathrm s. Find final position, displacement, distance travelled, average velocity and average speed.

Plan. Use signed leg lengths for displacement and positive lengths for distance.

rfinal=−2+9−4=3final coordinateΔr=3−(−2)=5change in positionD=9+4=13route lengthvˉ=5/5=1displacement rateaverage speed=13/5=2.6distance rate\begin{aligned} r_{\text{final}}&=-2+9-4=3 &&\text{final coordinate}\\ \Delta r&=3-(-2)=5 &&\text{change in position}\\ D&=9+4=13 &&\text{route length}\\ \bar v&=5/5=1 &&\text{displacement rate}\\ \text{average speed}&=13/5=2.6 &&\text{distance rate} \end{aligned}

Why subtract the initial coordinate? Final position includes the arbitrary choice of origin.

Answer. +3 m+3\,\mathrm m, +5 m+5\,\mathrm m, 13 m13\,\mathrm m, +1 m s−1+1\,\mathrm{m\,s^{-1}} and 2.6 m s−12.6\,\mathrm{m\,s^{-1}}, respectively.

Check. The distance from the origin is 3 m3\,\mathrm m, different from both travel totals.

Change only the return leg to 12 m12\,\mathrm m. Keep the five requested quantities and the elapsed time.

Worked example 2: travelling past the start

Question. A trolley starts at −2 m-2\,\mathrm m, moves 9 m9\,\mathrm m right and 12 m12\,\mathrm m left in 5 s5\,\mathrm s. Find final position, displacement, distance travelled, average velocity and average speed.

Plan. Recalculate the signed and unsigned totals.

rfinal=−2+9−12=−5final coordinateΔr=−5−(−2)=−3overall changeD=9+12=21all travelvˉ=−3/5=−0.6signed rateaverage speed=21/5=4.2non-negative rate\begin{aligned} r_{\text{final}}&=-2+9-12=-5 &&\text{final coordinate}\\ \Delta r&=-5-(-2)=-3 &&\text{overall change}\\ D&=9+12=21 &&\text{all travel}\\ \bar v&=-3/5=-0.6 &&\text{signed rate}\\ \text{average speed}&=21/5=4.2 &&\text{non-negative rate} \end{aligned}

Why can average velocity be negative? The trolley finishes left of its starting point.

Answer. −5 m-5\,\mathrm m, −3 m-3\,\mathrm m, 21 m21\,\mathrm m, −0.6 m s−1-0.6\,\mathrm{m\,s^{-1}} and 4.2 m s−14.2\,\mathrm{m\,s^{-1}}, respectively.

Check. Distance 2121 is greater than displacement magnitude 33.

An equation of motion links motion quantities on a stated interval; define its coordinate and positive direction before using it.

Watch out

Wrong: “The final position is −5-5, so displacement is −5-5.” Correct: displacement is −5−(−2)=−3 m-5-(-2)=-3\,\mathrm m. A starting coordinate contributes no extra travel.

Change the return leg again and track all five meanings.

Velocity, speed and acceleration signs

A velocity sign gives direction of motion; an acceleration sign gives direction of velocity change.

Signed rates and magnitudes

Velocity is rate of change of position. Speed is ∣v∣|v|. Acceleration is rate of change of velocity; its magnitude is ∣a∣|a|. Velocity and acceleration are vectors; speed and magnitude of acceleration are scalars.

For example, a velocity changing steadily from −8-8 to −2 m s−1-2\,\mathrm{m\,s^{-1}} in 3 s3\,\mathrm s has average acceleration +2 m s−2+2\,\mathrm{m\,s^{-2}}, found from (−2−(−8))/3({-2}-({-8}))/3. Its speed falls from 88 to 2 m s−12\,\mathrm{m\,s^{-1}}, despite positive acceleration. Swap the endpoint velocities and acceleration becomes −2 m s−2-2\,\mathrm{m\,s^{-2}}, making speed increase. Both acceleration magnitudes are 2 m s−22\,\mathrm{m\,s^{-2}}.

The average-acceleration rule, velocity change divided by elapsed time, is not in the formula sheet. At an instant with v≠0v\ne0, speed increases when vv and aa have the same sign and decreases when their signs differ. At v=0v=0, inspect the neighbouring motion: rest at an instant does not imply zero acceleration or a reversal. Constant velocity throughout an interval does imply zero acceleration.

Key idea

A negative acceleration can mean speeding up or slowing down. Its direction must be compared with the velocity direction.

Read the claim carefully: acceleration and its magnitude are different quantities.

Velocity from a displacement graph

The gradient of a position or displacement graph gives velocity; shifting the graph vertically leaves that velocity unchanged.

[DIAGRAM: asset_name: Piecewise displacement against time; asset_slug: displacement_segments; description: Displacement-time graph through (0,0), (2,6), (4,6) and (7,0), with time in seconds and displacement in metres; caption: Piecewise displacement against time]
Piecewise displacement against time

A rise of 6 m6\,\mathrm m over 2 s2\,\mathrm s gives 3 m s−13\,\mathrm{m\,s^{-1}}. The flat part means rest, not movement along a level road.

Reading a position gradient

On a straight segment,

v=ΔrΔtv=\frac{\Delta r}{\Delta t}

A chord gradient gives average velocity over an interval; a tangent gradient gives instantaneous velocity on a curve. The same rule holds for displacement-time graphs.

Not in the formula sheet: learn it.

For a sketch, label time and the vertical quantity with units. Straight segments represent constant velocity, horizontal segments rest, and curved segments changing velocity. A smooth maximum or minimum has zero instantaneous velocity. Ideal sharp corners stand for abrupt changes which real objects take time to make.

Worked example 3: from a displacement record

Question. A position-time graph joins (0,0),(2,6),(4,6),(7,0)(0,0),(2,6),(4,6),(7,0) with straight segments, in SI units. It has the same coordinates as the displayed displacement graph because the starting position is zero. Find the three velocities, overall displacement and distance travelled.

Plan. Take gradients, then compare signed and absolute position changes.

v1=6/2=3,v2=0first two slopesv3=(0−6)/3=−2return slopeΔr=0−0=0endpoint changeD=6+0+6=12all legs\begin{aligned} v_1&=6/2=3,\qquad v_2=0 &&\text{first two slopes}\\ v_3&=(0-6)/3=-2 &&\text{return slope}\\ \Delta r&=0-0=0 &&\text{endpoint change}\\ D&=6+0+6=12 &&\text{all legs} \end{aligned}

Why use slopes rather than heights? A height records displacement at that time.

Answer. Velocities 3,0,−2 m s−13,0,-2\,\mathrm{m\,s^{-1}}, displacement 0 m0\,\mathrm m, distance 12 m12\,\mathrm m.

Check. Signed travel 6−6=06-6=0 agrees with the graph's endpoints.

Change only the position graph's vertical offset by adding 5 m5\,\mathrm m to every coordinate.

Worked example 4: a different origin

Question. A position graph joins (0,5),(2,11),(4,11),(7,5)(0,5),(2,11),(4,11),(7,5) with straight segments, in SI units. Find the three velocities, overall displacement and distance travelled.

Plan. Form differences; a shared vertical offset cancels.

v1=(11−5)/2=3same risev2=(11−11)/2=0restv3=(5−11)/3=−2same fallΔr=5−5=0,D=6+6=12unchanged travel\begin{aligned} v_1&=(11-5)/2=3 &&\text{same rise}\\ v_2&=(11-11)/2=0 &&\text{rest}\\ v_3&=(5-11)/3=-2 &&\text{same fall}\\ \Delta r&=5-5=0,\qquad D=6+6=12 &&\text{unchanged travel} \end{aligned}

Why is displacement zero rather than 55? It is change in position, not the final coordinate.

Answer. 3,0,−2 m s−13,0,-2\,\mathrm{m\,s^{-1}}; 0 m0\,\mathrm m displacement; 12 m12\,\mathrm m distance.

Check. Subtracting 55 from the whole graph returns the first graph without changing any gradient.

Passing the origin and reversing direction are different events: one concerns graph height, the other a change in gradient sign.

Watch out

Wrong: “A position graph below the axis means negative velocity.” Correct: a downward gradient means negative velocity, wherever the graph lies.

Use a new offset to test whether the origin distracts you.

Acceleration and travel from velocity graphs

A velocity graph tells you acceleration through slope and displacement through signed area.

[DIAGRAM: asset_name: Positive and negative areas under a velocity graph; asset_slug: velocity_signed_areas; description: Velocity-time graph through (0,4), (2,4), (6,-4) and (8,-4), with its zero at t=4 and positive and negative travel regions shaded; caption: Positive and negative areas under a velocity graph]
Positive and negative areas under a velocity graph

The displayed positive travel is 12 m12\,\mathrm m, and negative travel contributes −12 m-12\,\mathrm m. The endpoint change is zero, although the journey length is 24 m24\,\mathrm m.

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