3.4 - Calculus in motion and practical models

3.4 - Calculus in motion and practical models

A changing position gives a velocity, and a changing velocity gives an acceleration. You will use differentiation for motion along a straight line, track distance after reversals and find rates or greatest values in simple practical models. Every model has its own units and valid interval.

Velocity and acceleration from displacement (Higher)

Can a moving object have zero acceleration? A model can answer precisely: its velocity might be changing at zero rate while the object is still travelling. Keep position, velocity and acceleration separate.

You already know

A derivative gives rate of change. Differentiate an integer-power expression term by term, then substitute a requested input. Negative graph gradient means decreasing output.

Displacement is signed position relative to a fixed origin and chosen positive direction. Positive velocity means increasing displacement; negative velocity means motion the other way. Speed is the magnitude ∣v∣|v|, so it is never negative.

Velocity and acceleration

v=dsdt,a=dvdtv=\frac{\mathrm{d}s}{\mathrm{d}t},\qquad a=\frac{\mathrm{d}v}{\mathrm{d}t}

For displacement in metres and time in seconds, velocity has units m s−1\mathrm{m\,s^{-1}} and acceleration m s−2\mathrm{m\,s^{-2}}. These relationships and differentiation rules are not in the Higher formulae sheet: learn them. This lesson is Higher only.

Differentiate displacement once for velocity, then differentiate velocity for acceleration. “Linear kinematics” means motion along a line; the displacement formula can still contain powers of time.

Worked example 1: moving with zero acceleration

Question. A particle has displacement s=t3−6t2+9ts=t^3-6t^2+9t metres for 0≤t≤40\leq t\leq4 seconds. Find its position, velocity, speed and acceleration at t=2t=2.

Plan. Differentiate twice in stages, then evaluate the appropriate expression for each quantity.

v=3t2−12t+9differentiate displacementa=6t−12differentiate velocitys(2)=8−24+18=2original positionv(2)=12−24+9=−3velocity at the timea(2)=12−12=0acceleration at the time\begin{aligned} v&=3t^2-12t+9 &&\text{differentiate displacement}\\ a&=6t-12 &&\text{differentiate velocity}\\ s(2)&=8-24+18=2 &&\text{original position}\\ v(2)&=12-24+9=-3 &&\text{velocity at the time}\\ a(2)&=12-12=0 &&\text{acceleration at the time} \end{aligned}

Why this step? Each expression describes a different quantity; zero acceleration does not set the velocity to zero.

Answer. Position 2 m2\,\mathrm m, velocity −3 m s−1-3\,\mathrm{m\,s^{-1}}, speed 3 m s−13\,\mathrm{m\,s^{-1}}, acceleration 0 m s−20\,\mathrm{m\,s^{-2}}.

Check. The object is on the positive side of the origin but moving in the negative direction, which is physically consistent.

Change only the requested time. The object can now be instantaneously at rest even though its acceleration is nonzero.

Worked example 2: rest with nonzero acceleration

Question. For the same displacement model, find position, velocity, speed and acceleration at t=1t=1.

Plan. Use the same expressions, evaluated at the new time.

s(1)=1−6+9=4positionv(1)=3−12+9=0velocitya(1)=6−12=−6accelerationspeed=∣0∣=0magnitude of velocity\begin{aligned} s(1)&=1-6+9=4 &&\text{position}\\ v(1)&=3-12+9=0 &&\text{velocity}\\ a(1)&=6-12=-6 &&\text{acceleration}\\ \text{speed}&=|0|=0 &&\text{magnitude of velocity} \end{aligned}

Why this step? Rest means zero velocity, while acceleration measures how velocity is changing through that instant.

Answer. Position 4 m4\,\mathrm m; velocity and speed 0 m s−10\,\mathrm{m\,s^{-1}}; acceleration −6 m s−2-6\,\mathrm{m\,s^{-2}}.

Check. Both times lie within the permitted interval. A turn in the position graph can have zero velocity but nonzero acceleration.

Acceleration's sign alone does not tell you whether speed increases. When velocity and acceleration have the same sign, speed increases locally; when they have opposite signs, speed decreases locally. At zero velocity, inspect the motion on either side instead.

Watch out

“a(2)=0a(2)=0, therefore the particle is at rest” is wrong. Calculate velocity: v(2)=−3 m s−1v(2)=-3\,\mathrm{m\,s^{-1}}, so it is moving. Rest requires v=0v=0.

Use another time in the same valid interval.

Rest, direction changes and distance (Higher)

A stop may reverse the motion, so subtracting the first position from the last can hide part of the journey. Find rest times from v=0v=0, then check the sign of velocity before and after each permitted time.

A change from positive to negative velocity reverses motion from increasing to decreasing displacement. A change from negative to positive reverses the other way. Zero velocity alone is not enough: for example, expand s=(t−1)3=t3−3t2+3t−1s=(t-1)^3=t^3-3t^2+3t-1, then differentiate to get v=3t2−6t+3=3(t−1)2v=3t^2-6t+3=3(t-1)^2. This is zero at t=1t=1 but positive on both sides, so there is no reversal.

Distance and displacement change

Displacement change is final position minus initial position. Distance travelled adds all lengths of the journey regardless of direction. Split at each reversal and add the absolute position changes. These relationships are not in the Higher formulae sheet: learn them.

For example, moving from position 00 to −4-4, then to 55 metres gives distance 4+9=13 m4+9=13\,\mathrm m, but displacement change 5 m5\,\mathrm m. Evaluate positions using the original displacement expression.

[DIAGRAM: asset_slug: m057_motion_distance; caption: Two direction reversals produce 12 metres of travel but only 4 metres of displacement change; description: Exact displacement-time graph s=t cubed minus 6t squared plus 9t for times 0 to 4 seconds, with rest at (1,4) and (3,0), alongside track arrows from 0 to 4, back to 0, then to 4 metres; file: m057_motion_distance.png; recommended_method: matplotlib]
Two direction reversals produce 12 metres of travel but only 4 metres of displacement change

The falling middle section means travel back towards the origin; the final rise adds distance again.

Worked example 3: distance after two reversals

Question. For s=t3−6t2+9ts=t^3-6t^2+9t metres, 0≤t≤40\leq t\leq4 seconds, find the rest times and total distance travelled.

Plan. Solve v=0v=0, check direction changes and add lengths between the start, reversals and end.

v=3t2−12t+9=3(t−1)(t−3)differentiate and factorisev=0 ⇒ t=1,3both within the intervals(0),s(1),s(3),s(4)=0,4,0,4evaluate positionsdistance=∣4−0∣+∣0−4∣+∣4−0∣add segment lengths=12 mtotal travel\begin{aligned} v&=3t^2-12t+9=3(t-1)(t-3) &&\text{differentiate and factorise}\\ v=0&\ \Rightarrow\ t=1,3 &&\text{both within the interval}\\ s(0),s(1),s(3),s(4)&=0,4,0,4 &&\text{evaluate positions}\\ \text{distance}&=|4-0|+|0-4|+|4-0| &&\text{add segment lengths}\\ &=12\,\mathrm m &&\text{total travel} \end{aligned}

Why this step? Velocity is positive before 11, negative between 11 and 33, then positive after 33, so both rest times reverse the motion.

Answer. Rest at 1 s1\,\mathrm s and 3 s3\,\mathrm s; distance 12 m12\,\mathrm m.

Check. The displacement change is 4 m4\,\mathrm m. The larger distance includes the return journey, as the diagram shows.

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