3.02f - Variable acceleration and calculus in motion

3.02f - Variable acceleration and calculus in motion

Differentiate and integrate to connect position, velocity and acceleration along a line. Use conditions to fix constants, check reversals and distinguish a signed displacement integral from total distance travelled.

Differentiate position and velocity

Can a particle be at rest without turning round? For velocity v=(t−2)2v=(t-2)^2, the particle stops for an instant at t=2t=2, yet moves positively on both sides. A zero on a velocity graph is only a candidate reversal.

You already know

Differentiate powers term by term; solve a factorised quadratic or use the quadratic formula. Velocity and acceleration signs decide whether speed increases or decreases. Check every root against the stated time interval.

Choose an origin and a positive direction. Write s(t)s(t) for signed position; displacement over an interval is the difference of two position values.

Instantaneous motion rates

v=dsdt,a=dvdt=d2sdt2v=\frac{\mathrm ds}{\mathrm dt},\qquad a=\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm d^2s}{\mathrm dt^2}

When ss is in metres and tt in seconds, vv has units m s−1\mathrm{m\,s^{-1}} and aa has units m s−2\mathrm{m\,s^{-2}}. These are tangent gradients, not interval averages.

Not in the formula sheet: learn it.

Set velocity, not acceleration, equal to zero to find rest.

Worked example 1: two reversals

Question. A particle has s=2t3−9t2+12ts=2t^3-9t^2+12t metres for 0≤t≤40\le t\le4 seconds. Find v(t)v(t), a(t)a(t) and all rest times, justifying whether any exist. Explain whether it speeds up or slows down at t=0.5t=0.5.

Plan. Differentiate twice, solve v=0v=0, then compare signs at the specified instant.

v=6t2−18t+12=6(t−1)(t−2)first derivativea=12t−18second derivativev=0 ⇒ t=1,2both in the intervalv(0.5)=4.5,a(0.5)=−12opposite signs\begin{aligned} v&=6t^2-18t+12=6(t-1)(t-2) &&\text{first derivative}\\ a&=12t-18 &&\text{second derivative}\\ v=0&\ \Rightarrow\ t=1,2 &&\text{both in the interval}\\ v(0.5)&=4.5,\qquad a(0.5)=-12 &&\text{opposite signs} \end{aligned}

Why compare signs? Positive velocity with negative acceleration has decreasing speed.

Answer. v=6t2−18t+12 m s−1v=6t^2-18t+12\,\mathrm{m\,s^{-1}}, a=12t−18 m s−2a=12t-18\,\mathrm{m\,s^{-2}}; rest at 11 and 2 s2\,\mathrm s. It slows at 0.5 s0.5\,\mathrm s.

Check. Velocity signs are positive, negative, positive on the three intervals, so both rest times are reversals. At t=1.5t=1.5, a=0a=0 but v=−1.5v=-1.5: zero acceleration is not rest.

Change only the cubic coefficient from 22 to 11. Keep the interval and every request.

Worked example 2: rejecting a late root

Question. A particle has s=t3−9t2+12ts=t^3-9t^2+12t metres for 0≤t≤40\le t\le4 seconds. Find v(t)v(t), a(t)a(t) and all rest times, justifying whether any exist. Explain whether it speeds up or slows down at t=0.5t=0.5.

Plan. Differentiate and solve, then retain only times in the model interval.

v=3t2−18t+12differentiate positiona=6t−18differentiate velocityt2−6t+4=0 ⇒ t=3±5solve v=03−5≈0.764,3+5≈5.24check intervalv(0.5)=3.75,a(0.5)=−15opposite signs\begin{aligned} v&=3t^2-18t+12 &&\text{differentiate position}\\ a&=6t-18 &&\text{differentiate velocity}\\ t^2-6t+4&=0\ \Rightarrow\ t=3\pm\sqrt5 &&\text{solve }v=0\\ 3-\sqrt5&\approx0.764,\qquad3+\sqrt5\approx5.24 &&\text{check interval}\\ v(0.5)&=3.75,\qquad a(0.5)=-15 &&\text{opposite signs} \end{aligned}

Why discard 3+53+\sqrt5? The model is given only up to 4 s4\,\mathrm s.

Answer. v=3t2−18t+12 m s−1v=3t^2-18t+12\,\mathrm{m\,s^{-1}}, a=6t−18 m s−2a=6t-18\,\mathrm{m\,s^{-2}}; the only rest time is (3−5) s(3-\sqrt5)\,\mathrm s. It slows at 0.5 s0.5\,\mathrm s.

Check. The accepted root changes velocity from positive to negative; the late root gives no conclusion about the specified journey.

A repeated velocity root such as v=(t−2)2v=(t-2)^2 does not change sign. At an endpoint, the given interval may also supply no motion on the other side.

Watch out

Wrong: “Every solution of v=0v=0 is a reversal.” Correct: first check the time interval, then check whether velocity changes sign.

Change only the cubic coefficient once more.

Integrate with initial conditions

An acceleration function tells you how velocity changes; it does not tell you the starting velocity or position.

Recovering motion from rates

For a known condition at time t0t_0,

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

Alternatively integrate indefinitely and determine each constant from its own condition. The acceleration integral has velocity units; the velocity integral has displacement units.

Not in the formula sheet: learn it.

The two constants fill different gaps. A condition at a non-zero time works in exactly the same way as an initial condition at zero.

Worked example 3: two non-zero starting values

Question. A particle has a=6t−4 m s−2a=6t-4\,\mathrm{m\,s^{-2}}, v(0)=2 m s−1v(0)=2\,\mathrm{m\,s^{-1}} and s(0)=5 ms(0)=5\,\mathrm m. Find its position function and its position and displacement from the start at t=2 st=2\,\mathrm s.

Plan. Integrate twice, using the velocity condition before integrating again.

v=3t2−4t+Cintegrate acceleration2=C ⇒ v=3t2−4t+2use v(0)s=t3−2t2+2t+Dintegrate full velocity5=D ⇒ s=t3−2t2+2t+5use s(0)s(2)=8−8+4+5=9positionΔs=9−5=4change from start\begin{aligned} v&=3t^2-4t+C &&\text{integrate acceleration}\\ 2&=C\ \Rightarrow\ v=3t^2-4t+2 &&\text{use }v(0)\\ s&=t^3-2t^2+2t+D &&\text{integrate full velocity}\\ 5&=D\ \Rightarrow\ s=t^3-2t^2+2t+5 &&\text{use }s(0)\\ s(2)&=8-8+4+5=9 &&\text{position}\\ \Delta s&=9-5=4 &&\text{change from start} \end{aligned}

Why use CC before the second integral? That constant contributes a linear position term.

Answer. s(t)=t3−2t2+2t+5s(t)=t^3-2t^2+2t+5 metres; position 9 m9\,\mathrm m, displacement 4 m4\,\mathrm m.

Check. Differentiating twice returns 6t−46t-4, and both stated conditions are satisfied.

Change only the velocity condition's time: use v(1)=2v(1)=2 instead of v(0)=2v(0)=2. Keep all requests.

Worked example 4: a condition away from zero

Question. A particle has a=6t−4 m s−2a=6t-4\,\mathrm{m\,s^{-2}}, v(1)=2 m s−1v(1)=2\,\mathrm{m\,s^{-1}} and s(0)=5 ms(0)=5\,\mathrm m. Find its position function and its position and displacement from the start at t=2 st=2\,\mathrm s.

Plan. Substitute the actual condition time into the velocity antiderivative.

v=3t2−4t+Cintegrate acceleration2=3−4+C ⇒ C=3condition at 1s=t3−2t2+3t+Dintegrate velocityD=5position at 0s(2)=8−8+6+5=11positionΔs=11−5=6displacement\begin{aligned} v&=3t^2-4t+C &&\text{integrate acceleration}\\ 2&=3-4+C\ \Rightarrow\ C=3 &&\text{condition at }1\\ s&=t^3-2t^2+3t+D &&\text{integrate velocity}\\ D&=5 &&\text{position at }0\\ s(2)&=8-8+6+5=11 &&\text{position}\\ \Delta s&=11-5=6 &&\text{displacement} \end{aligned}

Why is CC not 22? The known velocity occurs at time 11, when the other terms do not vanish.

Answer. s(t)=t3−2t2+3t+5s(t)=t^3-2t^2+3t+5 metres; position 11 m11\,\mathrm m, displacement 6 m6\,\mathrm m.

Check. v(1)=3−4+3=2v(1)=3-4+3=2 and s(0)=5s(0)=5.

Do not add a second arbitrary constant if a definite integral has already used the relevant condition.

Watch out

Wrong: “Integrating twice gives one constant.” Correct: each integration has its own missing information. A velocity condition fixes the velocity constant; a position condition fixes the position constant.

Move the velocity condition to another time.

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