Acceleration — Question 3

Question 3

Differentiate position twice, then compare signs

On an east-positive straight line, a point has the exact position history $$x(t)=1+8t-3t^2+\tfrac13t^3,\qquad0\leq t\leq5.$$ Here $t$ is numerical time in seconds and $x$ numerical position in metres. Differentiate term by term using $d(ct^n)/dt=nc t^{n-1}$: $v=dx/dt$ in m/s and $a=dv/dt$ in $\mathrm{m/s^2}$.

1. Find $v(t)$ in factored form and find $a(t)$. Identify all interior times when velocity is zero and when acceleration is zero.

2. Make a sign table for velocity and acceleration on the intervals separated by those times. Determine direction of motion and whether speed increases or decreases on each interval.

3. At 3 s, find velocity, speed, and acceleration. Does zero acceleration imply zero velocity or constant velocity over a surrounding interval?

4. At each zero-velocity instant, give the acceleration and determine whether a reversal occurs. Explain why zero velocity does not force zero acceleration.

5. Find the average acceleration over 0–5 s and compare it with the instantaneous acceleration at each endpoint, using the polynomial’s smooth extension to define endpoint derivatives.

Question 3 - Solution

Differentiate position twice, then compare signs

(a) Velocity and acceleration are different derivatives.

$$\boxed{v(t)=t^2-6t+8=(t-2)(t-4)\ \mathrm{m/s}},\qquad \boxed{a(t)=2t-6\ \mathrm{m/s^2}}.$$ Velocity is zero at $\boxed{2,4\ \mathrm{s}}$; acceleration is zero at $\boxed{3\ \mathrm{s}}$. The formulas use the problem’s numerical-time convention.

(b) Use both signs to decide the speed change.

$$\frac{d(v^2)}{dt}=2va;\quad va>0\Rightarrow\text{speed increases},\quad va<0\Rightarrow\text{speed decreases}.$$

$$\begin{aligned}0<t<2 &: (v,a)=(+,-),&&\text{east, slowing},\\2<t<3 &: (v,a)=(-,-),&&\text{west, speeding up},\\3<t<4 &: (v,a)=(-,+),&&\text{west, slowing},\\4<t<5 &: (v,a)=(+,+),&&\text{east, speeding up}.\end{aligned}$$ When $v>0$, speed equals $v$; when $v<0$, speed equals $-v$. Thus the same acceleration sign can accompany either increasing or decreasing speed.

(c) A zero derivative at one instant is not a constant history.

At 3 s, $\boxed{v=-1\ \mathrm{m/s},\ s=1\ \mathrm{m/s},\ a=0\ \mathrm{m/s^2}}$. The point is moving west. Its velocity varies on every surrounding interval; only its derivative at the selected instant is zero. Zero acceleration over an entire interval would be a stronger condition.

(d) A value can vanish while its derivative does not.

At 2 s, $a=-2\ \mathrm{m/s^2}$ and velocity changes from positive to negative. At 4 s, $a=+2\ \mathrm{m/s^2}$ and velocity changes from negative to positive. Both are reversals with nonzero acceleration. Being instantaneously at rest constrains $v$, not necessarily its derivative.

(e) An interval ratio need not equal either endpoint value.

Since $v(0)=8$ and $v(5)=3\ \mathrm{m/s}$, $$\boxed{\bar a=(3-8)/5=-1\ \mathrm{m/s^2}}.$$ The smooth polynomial gives $a(0)=-6$ and $a(5)=+4\ \mathrm{m/s^2}$. The average describes the net velocity change per elapsed time, not either endpoint slope.