Mk5, Mk6 - Variable acceleration and calculus in motion

Mk5, Mk6 - Variable acceleration and calculus in motion

Connect position, velocity and acceleration by differentiating and integrating with respect to time. Recover motion from conditions, distinguish signed change from total travel, and justify predictions from a variable-acceleration model.

Differentiate position and velocity

Must a particle turn whenever it stops? The position r=(t−2)3r=(t-2)^3 has velocity 3(t−2)23(t-2)^2: it moves positively, pauses at t=2t=2, then continues positively. A stationary instant need not split the journey into opposite directions.

You already know

Differentiate powers and exponentials; integrate term by term. Solve exponential equations using natural logarithms. Position is signed, speed is ∣v∣|v|, and a velocity zero must be checked against its time interval and neighbouring signs.

Use r(t)r(t) for position relative to a fixed origin, with a stated positive direction. A position difference is displacement; the position itself need not start at zero.

Rates in one-dimensional motion

v=drdt,a=dvdt=d2rdt2v=\frac{\mathrm dr}{\mathrm dt},\qquad a=\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm d^2r}{\mathrm dt^2}

Position in metres and time in seconds give velocity in m s−1\mathrm{m\,s^{-1}} and acceleration in m s−2\mathrm{m\,s^{-2}}. Velocity is a position-time tangent gradient; acceleration is a velocity-time tangent gradient.

Not in the formula sheet: learn it.

The constant term in a position model sets its offset but disappears on differentiation.

Worked example 1: a smooth speed increase

Question. For 0≤t≤30\le t\le3 seconds, a particle has position r=5+12t+8(e−t−1)r=5+12t+8(\mathrm e^{-t}-1) metres. Determine velocity, acceleration and any rest times, and whether speed initially increases or decreases.

Plan. Differentiate twice, solve v=0v=0, then compare the starting signs.

v=12−8e−tdifferentiate positiona=8e−tdifferentiate velocityv=0 ⇒ e−t=32rest equation0<e−t≤1(t≥0)root impossible in intervalv(0)=4,a(0)=8same positive signs\begin{aligned} v&=12-8\mathrm e^{-t} &&\text{differentiate position}\\ a&=8\mathrm e^{-t} &&\text{differentiate velocity}\\ v=0&\ \Rightarrow\ \mathrm e^{-t}=\tfrac32 &&\text{rest equation}\\ 0<\mathrm e^{-t}&\le1\quad(t\ge0) &&\text{root impossible in interval}\\ v(0)&=4,\qquad a(0)=8 &&\text{same positive signs} \end{aligned}

Why reject the rest equation? It demands an exponential larger than 11 at a non-negative time.

Answer. v=12−8e−t m s−1v=12-8\mathrm e^{-t}\,\mathrm{m\,s^{-1}}, a=8e−t m s−2a=8\mathrm e^{-t}\,\mathrm{m\,s^{-2}}; no rest times. Speed initially increases.

Check. r(0)=5r(0)=5; acceleration is positive while velocity increases from 44 towards 12 m s−112\,\mathrm{m\,s^{-1}}.

Change only the coefficient of tt from 1212 to 44; keep the interval and every request.

Worked example 2: an exponential reversal

Question. For 0≤t≤30\le t\le3, a particle has position r=5+4t+8(e−t−1)r=5+4t+8(\mathrm e^{-t}-1) metres. Determine velocity, acceleration and any rest times, and whether speed initially increases or decreases.

Plan. Use the new velocity expression and solve its zero exactly.

v=4−8e−t,a=8e−tdifferentiate twice4−8e−t=0 ⇒ e−t=12rest conditiont=ln⁡2take logarithmsv(0)=−4,a(0)=8opposite signs\begin{aligned} v&=4-8\mathrm e^{-t},\qquad a=8\mathrm e^{-t} &&\text{differentiate twice}\\ 4-8\mathrm e^{-t}&=0 \ \Rightarrow\ \mathrm e^{-t}=\tfrac12 &&\text{rest condition}\\ t&=\ln2 &&\text{take logarithms}\\ v(0)&=-4,\qquad a(0)=8 &&\text{opposite signs} \end{aligned}

Why does positive acceleration initially slow the particle? It makes a negative velocity less negative.

Answer. v=4−8e−t m s−1v=4-8\mathrm e^{-t}\,\mathrm{m\,s^{-1}}, a=8e−t m s−2a=8\mathrm e^{-t}\,\mathrm{m\,s^{-2}}; rest at t=ln⁡2t=\ln2 seconds, within the interval. Speed initially decreases.

Check. Velocity increases through zero at ln⁡2\ln2, so this rest time is a reversal, unlike the opening cubic example.

A repeated velocity root can leave its sign unchanged, and an endpoint zero supplies no automatic evidence about motion outside the model.

Watch out

Wrong: “Set acceleration to zero to find rest.” Correct: rest means v=0v=0. A zero of acceleration concerns a stationary value of velocity, not automatically a stationary particle.

Change the linear coefficient once more.

Integrate with initial conditions

Integration gives a family of motions; the supplied conditions select the particular particle.

Integrating rates with conditions

v(t)=v(t0)+∫t0ta(τ) dτr(t)=r(t0)+∫t0tv(τ) dτ\begin{aligned} v(t)&=v(t_0)+\int_{t_0}^{t}a(\tau)\,\mathrm d\tau\\ r(t)&=r(t_0)+\int_{t_0}^{t}v(\tau)\,\mathrm d\tau \end{aligned}

An indefinite-integral route uses one constant for velocity and another for position. Acceleration-time area gives velocity change; velocity-time area gives position change.

Not in the formula sheet: learn it.

A condition at time 11 or 22 is not an initial condition at zero. Substitute the stated time rather than assigning its value directly to a constant.

Worked example 3: integrate twice

Question. A particle has a=4t+2 m s−2a=4t+2\,\mathrm{m\,s^{-2}}, v(0)=−3 m s−1v(0)=-3\,\mathrm{m\,s^{-1}} and r(0)=7 mr(0)=7\,\mathrm m. Determine r(t)r(t), position at t=3t=3 and displacement during the first three seconds.

Plan. Fix the velocity constant, integrate the complete velocity, then fix the position constant.

v=2t2+2t+Cintegrate accelerationC=−3use v(0)r=23t3+t2−3t+Dintegrate velocityD=7use r(0)r(3)=18+9−9+7=25positionΔr=25−7=18displacement\begin{aligned} v&=2t^2+2t+C &&\text{integrate acceleration}\\ C&=-3 &&\text{use }v(0)\\ r&=\tfrac23t^3+t^2-3t+D &&\text{integrate velocity}\\ D&=7 &&\text{use }r(0)\\ r(3)&=18+9-9+7=25 &&\text{position}\\ \Delta r&=25-7=18 &&\text{displacement} \end{aligned}

Why keep the constant −3-3 during the second integration? The initial velocity contributes −3t-3t to position.

Answer. r(t)=23t3+t2−3t+7r(t)=\frac23t^3+t^2-3t+7 metres; position 25 m25\,\mathrm m, displacement 18 m18\,\mathrm m.

Check. Differentiating twice returns a=4t+2a=4t+2, and both zero-time conditions hold.

Change only the time of the velocity condition: replace v(0)=−3v(0)=-3 by v(1)=−3v(1)=-3.

Worked example 4: use the actual condition time

Question. A particle has a=4t+2 m s−2a=4t+2\,\mathrm{m\,s^{-2}}, v(1)=−3 m s−1v(1)=-3\,\mathrm{m\,s^{-1}} and r(0)=7 mr(0)=7\,\mathrm m. Determine r(t)r(t), position at t=3t=3 and displacement during the first three seconds.

Plan. Put t=1t=1 into the velocity antiderivative before integrating again.

v=2t2+2t+Cintegrate−3=2+2+C ⇒ C=−7condition at 1r=23t3+t2−7t+Dintegrate complete velocityD=7position conditionr(3)=18+9−21+7=13positionΔr=13−7=6signed change\begin{aligned} v&=2t^2+2t+C &&\text{integrate}\\ -3&=2+2+C\ \Rightarrow\ C=-7 &&\text{condition at }1\\ r&=\tfrac23t^3+t^2-7t+D &&\text{integrate complete velocity}\\ D&=7 &&\text{position condition}\\ r(3)&=18+9-21+7=13 &&\text{position}\\ \Delta r&=13-7=6 &&\text{signed change} \end{aligned}

Why is the constant −7-7, not −3-3? At t=1t=1, the other velocity terms already contribute 44.

Answer. r(t)=23t3+t2−7t+7r(t)=\frac23t^3+t^2-7t+7 metres; position 13 m13\,\mathrm m, displacement 6 m6\,\mathrm m.

Check. v(1)=2+2−7=−3v(1)=2+2-7=-3, and r(0)=7r(0)=7.

A definite integral which already uses a condition needs no additional arbitrary constant.

Watch out

Wrong: “Position at time 33 is the integral from 00 to 33.” Correct: the integral gives displacement. Add r(0)r(0) for the position.

Shift the velocity condition again, keeping the other data.

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