4.2 - Motion with calculus

4.2 - Motion with calculus

When acceleration changes with time, a single constant-acceleration equation no longer describes the whole motion. You will differentiate position to obtain velocity and acceleration, integrate rates using initial conditions, and split a journey at genuine direction changes to find total distance. The calculations use the power methods already developed in Pure Mathematics 1.

Differentiating displacement

Can a particle be at rest for an instant while its acceleration is nonzero? A position function answers both questions separately: its first derivative gives velocity and its second gives acceleration.

Let s(t)s(t) be signed displacement from a fixed origin, in metres, at time tt seconds. A small change in position divided by the corresponding time change approximates velocity; the tangent gradient is the instantaneous velocity.

Rates of motion

v=dsdt,a=dvdt=d2sdt2.v=\frac{\mathrm ds}{\mathrm dt},\qquad a=\frac{\mathrm dv}{\mathrm dt}=\frac{\mathrm d^2s}{\mathrm dt^2}.

Speed is ∣v∣|v|. Differentiate only where the stated function is defined and differentiable; units are m s−1\mathrm{m\,s^{-1}} for velocity and m s−2\mathrm{m\,s^{-2}} for acceleration.

Not in the formula booklet: learn it.

P1's power rule applies directly: the derivative of tnt^n is ntn−1nt^{n-1}, with constants differentiating to zero. Keep signs when finding velocity; take its magnitude only when speed is requested.

Worked example 1: quadratic position

Question. A particle has displacement s=2t2−8t+3s=2t^2-8t+3 metres for t≥0t\ge0. Find velocity and acceleration, then find its speed at t=1t=1.

Plan. Differentiate twice, evaluate velocity, then take its magnitude for speed.

v=4t−8first derivativea=4second derivativev(1)=4−8=−4 m s−1\begin{aligned} v&=4t-8 &&\text{first derivative}\\ a&=4 &&\text{second derivative}\\ v(1)&=4-8=-4\,\mathrm{m\,s^{-1}} \end{aligned}

Answer. v=4t−8 m s−1v=4t-8\,\mathrm{m\,s^{-1}}, a=4 m s−2a=4\,\mathrm{m\,s^{-2}}; speed at t=1t=1 is 4 m s−14\,\mathrm{m\,s^{-1}}, with motion in the negative direction.

Check. A small symmetric interval gives [s(1.1)−s(0.9)]/0.2=−4 m s−1[s(1.1)-s(0.9)]/0.2=-4\,\mathrm{m\,s^{-1}}, agreeing with the tangent gradient. Acceleration is constant here.

Change just the power in the first term from t2t^2 to t3t^3. Acceleration will now depend on time.

Worked example 2: cubic position

Question. A particle has displacement s=2t3−8t+3s=2t^3-8t+3 metres for t≥0t\ge0. Find velocity and acceleration, then find its speed at t=1t=1.

Plan. Apply the same derivative chain and keep the two rates distinct.

v=6t2−8a=12tv(1)=6−8=−2 m s−1a(1)=12 m s−2\begin{aligned} v&=6t^2-8\\ a&=12t\\ v(1)&=6-8=-2\,\mathrm{m\,s^{-1}}\\ a(1)&=12\,\mathrm{m\,s^{-2}} \end{aligned}

Answer. v=6t2−8 m s−1v=6t^2-8\,\mathrm{m\,s^{-1}}, a=12t m s−2a=12t\,\mathrm{m\,s^{-2}}; speed at t=1t=1 is 2 m s−12\,\mathrm{m\,s^{-1}}.

Check. The average change of velocity between t=0.9t=0.9 and t=1.1t=1.1 is [v(1.1)−v(0.9)]/0.2=12 m s−2[v(1.1)-v(0.9)]/0.2=12\,\mathrm{m\,s^{-2}}, confirming the acceleration at the midpoint for this quadratic velocity. Since aa varies, the constant-acceleration equations cannot be used over a finite interval here.

Adding a constant to ss changes the origin of position but leaves vv and aa unchanged. A zero of ss means crossing the position origin; a zero of vv means an instantaneous stop. They answer different questions.

Watch out

To find an instant of rest, solve v=0v=0, not s=0s=0 or a=0a=0. A zero acceleration identifies zero instantaneous change of velocity, which need not mean rest.

Differentiate the position before substituting the time.

Integrating a rate

Integration reverses the derivative chain. From acceleration you obtain velocity, and from velocity you obtain position. Each integration introduces a separate constant; use the information appropriate to that quantity to determine it.

Recovering motion functions

v=∫a dt,s=∫v dt.v=\int a\,\mathrm dt,\qquad s=\int v\,\mathrm dt.

Each indefinite integral includes its own arbitrary constant. Use different names, such as CC and DD: a specified velocity fixes the first, and a specified position fixes the second.

These motion relationships are not in the formula booklet: learn them.

Use the P1 power method: increase the power by one and divide by the new power. For n≠−1n\ne-1, an antiderivative of tnt^n is tn+1/(n+1)t^{n+1}/(n+1). Differentiate your result to check it.

Worked example 3: integrating twice with initial conditions

Question. A particle has acceleration a=6t−4 m s−2a=6t-4\,\mathrm{m\,s^{-2}}. At t=0t=0, its velocity is 3 m s−13\,\mathrm{m\,s^{-1}} and its displacement is 5 m5\,\mathrm{m}. Find v(t)v(t) and s(t)s(t).

Plan. Integrate acceleration, use the velocity condition, then integrate that velocity and use the position condition.

v=3t2−4t+Cv(0)=3⟹C=3s=t3−2t2+3t+Ds(0)=5⟹D=5\begin{aligned} v&=3t^2-4t+C\\ v(0)=3&\quad\Longrightarrow\quad C=3\\ s&=t^3-2t^2+3t+D\\ s(0)=5&\quad\Longrightarrow\quad D=5 \end{aligned}

Answer. v=3t2−4t+3 m s−1v=3t^2-4t+3\,\mathrm{m\,s^{-1}}, s=t3−2t2+3t+5 ms=t^3-2t^2+3t+5\,\mathrm{m}.

Check. Differentiating ss gives the stated vv; differentiating vv gives 6t−46t-4. At zero, the functions give v=3v=3 and s=5s=5, satisfying both conditions.

Change only the initial velocity to −3 m s−1-3\,\mathrm{m\,s^{-1}}. The acceleration is identical, but its motion is not.

Worked example 4: a different initial velocity

Question. A particle again has acceleration a=6t−4 m s−2a=6t-4\,\mathrm{m\,s^{-2}}. At t=0t=0, its velocity is −3 m s−1-3\,\mathrm{m\,s^{-1}} and displacement is 5 m5\,\mathrm{m}. Find v(t)v(t) and s(t)s(t).

Plan. Find the integration constants from these new conditions.

v=3t2−4t+C,C=−3s=t3−2t2−3t+D,D=5\begin{aligned} v&=3t^2-4t+C,\qquad C=-3\\ s&=t^3-2t^2-3t+D,\qquad D=5 \end{aligned}

Answer. v=3t2−4t−3 m s−1v=3t^2-4t-3\,\mathrm{m\,s^{-1}}, s=t3−2t2−3t+5 ms=t^3-2t^2-3t+5\,\mathrm{m}.

Check. The two velocities differ by −6-6 at every time. Their positions differ by −6t-6t: the same acceleration preserves the initial velocity difference. Both particles still start at position 55.

A condition can be given at a time other than zero. For example, if v=3t2−4t+Cv=3t^2-4t+C and v(2)=7v(2)=7, then 7=12−8+C7=12-8+C, giving C=3C=3. Substitute the stated time, rather than setting the constant equal to the given velocity automatically.

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