Q4 - Calculus for straight-line motion

Q4 - Calculus for straight-line motion

Use differentiation and integration when acceleration varies, determine constants from conditions at stated times, and calculate total distance by finding reversals. You will also check whether a mathematical motion model supports the physical conclusions drawn from it.

Velocity and acceleration as derivatives

Can a particle have zero acceleration while still moving backwards? The position model r=t3−6t2+9t+2r=t^3-6t^2+9t+2 does exactly that at t=2t=2. Calculus lets us examine one instant without pretending that acceleration remains constant throughout the journey.

You already know

Velocity is signed: a negative value means motion in the chosen negative direction. Speed is ∣v∣|v|. Differentiation finds an instantaneous rate of change; a constant differentiates to zero.

Let r(t)r(t) be signed position along a line. Displacement from the initial position is s(t)=r(t)−r(0)s(t)=r(t)-r(0). The constant r(0)r(0) disappears on differentiation, so position and displacement have the same time derivatives.

Kinematic derivatives

v=drdt=r˙a=dvdt=d2rdt2=r¨v=\frac{dr}{dt}=\dot r \qquad a=\frac{dv}{dt}=\frac{d^2r}{dt^2}=\ddot r

A dot denotes differentiation with respect to time. These relationships apply whether or not acceleration is constant. Not in the formula booklet: learn it.

Position in metres differentiates to velocity in m s−1\mathrm{m\,s^{-1}}, then acceleration in m s−2\mathrm{m\,s^{-2}}. Rest is a velocity condition, v=0v=0; zero acceleration means the velocity is momentarily not changing.

Worked example 1: zero acceleration without rest

Question. For 0≤t≤40\le t\le4, a particle has position r=t3−6t2+9t+2r=t^3-6t^2+9t+2 metres, where tt is in seconds. Find velocity and acceleration, and describe its motion at t=2t=2.

Plan. Differentiate twice, then evaluate both rates at the requested time.

v=3t2−12t+9=3(t−1)(t−3)differentiate positiona=6t−12differentiate velocityv(2)=12−24+9=−3velocity at the instanta(2)=12−12=0acceleration at the instant\begin{aligned} v&=3t^2-12t+9=3(t-1)(t-3) && \text{differentiate position} \\ a&=6t-12 && \text{differentiate velocity} \\ v(2)&=12-24+9=-3 && \text{velocity at the instant} \\ a(2)&=12-12=0 && \text{acceleration at the instant} \end{aligned}

Why calculate both? Acceleration alone cannot tell you whether the particle is moving or at rest.

Answer. At t=2t=2, velocity is −3 m s−1-3\,\mathrm{m\,s^{-1}}, speed is 3 m s−13\,\mathrm{m\,s^{-1}} and acceleration is zero. The particle is moving in the negative direction.

Check. The rest times are t=1t=1 and t=3t=3, found from v=0v=0, so t=2t=2 is not a rest time.

Worked example 2 changes the position's function family. The same two derivatives now require product and chain rules, both of which can appear in A-level motion questions.

Worked example 2: a damped trigonometric position

Question. A particle has signed position r=e−tsin⁡2tr=e^{-t}\sin2t metres for t≥0t\ge0, with time in seconds and trigonometric arguments in radians. Find velocity and acceleration, including their initial values.

Plan. Use the product and chain rules for each differentiation, keeping the exponential factor outside.

v=−e−tsin⁡2t+2e−tcos⁡2tdifferentiate both factors=e−t(2cos⁡2t−sin⁡2t)a=−e−t(2cos⁡2t−sin⁡2t)+e−t(−4sin⁡2t−2cos⁡2t)product rule again=e−t(−3sin⁡2t−4cos⁡2t)collect terms\begin{aligned} v&=-e^{-t}\sin2t+2e^{-t}\cos2t && \text{differentiate both factors} \\ &=e^{-t}(2\cos2t-\sin2t) \\ a&=-e^{-t}(2\cos2t-\sin2t) \\ &\quad+e^{-t}(-4\sin2t-2\cos2t) && \text{product rule again} \\ &=e^{-t}(-3\sin2t-4\cos2t) && \text{collect terms} \end{aligned}

Why keep the exponential factor? It makes the two contributions to each sine and cosine coefficient easy to combine.

Answer. v=e−t(2cos⁡2t−sin⁡2t) m s−1v=e^{-t}(2\cos2t-\sin2t)\,\mathrm{m\,s^{-1}} and a=e−t(−3sin⁡2t−4cos⁡2t) m s−2a=e^{-t}(-3\sin2t-4\cos2t)\,\mathrm{m\,s^{-2}}. Initially, v(0)=2 m s−1v(0)=2\,\mathrm{m\,s^{-1}} and a(0)=−4 m s−2a(0)=-4\,\mathrm{m\,s^{-2}}.

Check. Differentiating the expanded velocity gives cosine coefficient −2−2=−4-2-2=-4 and sine coefficient 1−4=−31-4=-3, agreeing with the result.

The coefficients in a physical expression carry whatever units are needed for consistent dimensions. Trigonometric arguments are in radians when using these calculus rules. The elementary derivatives, product rule and chain rule must be recalled; they are not supplied in the booklet.

Watch out

Wrong: ddt(p+2t−qe−0.2t)=1+2+0.2qe−0.2t\frac{d}{dt}(p+2t-qe^{-0.2t})=1+2+0.2qe^{-0.2t} for constants p,qp,q. The constant pp differentiates to zero. Right: v=2+0.2qe−0.2tv=2+0.2qe^{-0.2t}, then a=−0.04qe−0.2ta=-0.04qe^{-0.2t}; each chain-rule factor matters.

Try a new product. The final numerical speed needs the magnitude of the signed velocity.

Integration and initial conditions

Integration reverses the rate relationships, but each integration introduces a constant determined by the matching condition.

If acceleration is 6t−46t-4, velocity must have the form 3t2−4t+C3t^2-4t+C. Different choices of CC describe different initial velocities, all with the same acceleration. A position condition cannot replace the velocity condition.

Kinematic integrals

v(t)=∫a(t) dtr(t)=∫v(t) dtv(t)=\int a(t)\,dt \qquad r(t)=\int v(t)\,dt

Include a separate constant for each integration, then apply the condition at its stated time. Also,

r(t2)−r(t1)=∫t1t2v(t) dtr(t_2)-r(t_1)=\int_{t_1}^{t_2}v(t)\,dt

Not in the formula booklet: learn it.

“Initially” means t=0t=0. A condition given at t=2t=2 must instead be substituted at 22. For example, integrating v=2t+1v=2t+1 gives r=t2+t+Cr=t^2+t+C; if r(2)=7r(2)=7, then 7=4+2+C7=4+2+C, so C=1C=1. Assuming initial position zero would describe a different journey.

Worked example 3: two integrations and two conditions

Question. A particle has a=6t−4 m s−2a=6t-4\,\mathrm{m\,s^{-2}}, v(0)=3 m s−1v(0)=3\,\mathrm{m\,s^{-1}} and r(0)=−2 mr(0)=-2\,\mathrm m. Find velocity and position as functions of time.

Plan. Integrate acceleration and apply the velocity condition before integrating again and applying the position condition.

v=3t2−4t+Cintegrate acceleration3=Cuse v(0)=3r=t3−2t2+3t+Dintegrate the complete velocity−2=Duse r(0)=−2\begin{aligned} v&=3t^2-4t+C && \text{integrate acceleration} \\ 3&=C && \text{use }v(0)=3 \\ r&=t^3-2t^2+3t+D && \text{integrate the complete velocity} \\ -2&=D && \text{use }r(0)=-2 \end{aligned}

Why find CC before integrating again? The constant velocity contribution produces a CtCt term in position, so it affects the whole later journey.

Answer. v=3t2−4t+3 m s−1v=3t^2-4t+3\,\mathrm{m\,s^{-1}} and r=t3−2t2+3t−2r=t^3-2t^2+3t-2 metres.

Check. Differentiating position recovers velocity; differentiating velocity recovers 6t−46t-4. The two values at zero are 33 and −2-2, as required.

Worked example 4 changes the acceleration family to an exponential. The bookkeeping is unchanged, but reverse-chain signs must be checked carefully.

Worked example 4: an exponential acceleration

Question. A particle has a=2e−t m s−2a=2e^{-t}\,\mathrm{m\,s^{-2}}, v(0)=1 m s−1v(0)=1\,\mathrm{m\,s^{-1}} and r(0)=4 mr(0)=4\,\mathrm m. Find velocity and position.

Plan. Integrate the exponential twice, applying the corresponding condition after each integration.

v=−2e−t+Cintegrate 2e−t1=−2+Cuse initial velocityv=3−2e−tr=3t+2e−t+Dintegrate velocity4=2+Duse initial positionr=2+3t+2e−t\begin{aligned} v&=-2e^{-t}+C && \text{integrate }2e^{-t} \\ 1&=-2+C && \text{use initial velocity} \\ v&=3-2e^{-t} \\ r&=3t+2e^{-t}+D && \text{integrate velocity} \\ 4&=2+D && \text{use initial position} \\ r&=2+3t+2e^{-t} \end{aligned}

Why is the first antiderivative negative? Differentiating −2e−t-2e^{-t} supplies two negative signs, giving the positive acceleration 2e−t2e^{-t}.

Answer. v=3−2e−t m s−1v=3-2e^{-t}\,\mathrm{m\,s^{-1}} and r=2+3t+2e−tr=2+3t+2e^{-t} metres.

Check. At zero, v=3−2=1v=3-2=1 and r=2+2=4r=2+2=4; differentiating twice recovers the given acceleration.

A definite integral is often the shortest route to displacement over an interval because the position constant cancels. Elementary power, exponential and trigonometric integrals are recall results, rather than separately supplied kinematic formulae.

At A-level, a motion question can require any appropriate Pure Mathematics calculus technique. For example, a polynomial times an exponential may need integration by parts. Applying the supplied integration-by-parts formula twice gives

∫t2e−t dt=−t2e−t+2∫te−t dtdifferentiate the polynomial=−t2e−t+2(−te−t−e−t)+Cparts once more=−(t2+2t+2)e−t+C\begin{aligned} \int t^2e^{-t}\,dt&=-t^2e^{-t}+2\int te^{-t}\,dt && \text{differentiate the polynomial} \\ &=-t^2e^{-t}+2(-te^{-t}-e^{-t})+C && \text{parts once more} \\ &=-(t^2+2t+2)e^{-t}+C \end{aligned}

This derived antiderivative is not separately supplied in the booklet. If a=t2e−ta=t^2e^{-t}, it is the velocity family; determine CC from the velocity condition just as before. Substitution and partial fractions also transfer from Pure Mathematics when the function calls for them. The motion interpretation changes neither the calculus rules nor the need for constants.

Watch out

Wrong: r=t3−2t2+3t+3r=t^3-2t^2+3t+3, reusing the velocity constant as the position constant in Worked example 3. This gives r(0)=3r(0)=3, contradicting −2-2. Right: use separate symbols, finding C=3C=3 from velocity and D=−2D=-2 from position; the constants also have different units.

The next acceleration uses a trigonometric function. Apply the same two-condition method, with the appropriate elementary integrals.

Total distance when direction changes

An integral of velocity gives signed displacement, so total distance needs the velocity sign on the whole interval.

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