Instantaneous Velocity and Speed — Question 3

Question 3

Find reversals before adding traveled lengths

In this problem, $t$ denotes numerical time in seconds and $x$ numerical position in metres. A point on an east-positive line has the complete history $$x(t)=t^3-6t^2+9t=t(t-3)^2,\qquad0\leq t\leq4.$$ For a polynomial, differentiate term by term using $d(ct^n)/dt=nc t^{n-1}$; a constant has derivative zero. The derivative gives velocity in m/s. Speed is $|v|$.

1. Find $v(t)$ in factored form and identify every interior instant with zero speed.

2. Determine the direction of motion on the intervals separated by those instants. Which zero-speed instants are reversals? Justify using signs on both sides.

3. Find position, velocity, and speed at 2 s. Explain why the positive position does not fix the sign of velocity.

4. Find total distance and displacement over 0–4 s, then both full-interval averages. Use turning positions, without integrating velocity. Why would taking only the endpoint separation miss traveled distance?

Question 3 - Solution

Find reversals before adding traveled lengths

(a) Differentiate, then factor to expose the zeros.

$$\boxed{v(t)=3t^2-12t+9=3(t-1)(t-3)\ \mathrm{m/s}}.$$ Here and below $t$ is the numerical time in seconds. Speed is zero at $\boxed{t=1\ \mathrm{s},\ 3\ \mathrm{s}}$.

(b) Test signs between the zeros.

$$\operatorname{sgn}v=\begin{cases}+1,&0<t<1,\\-1,&1<t<3,\\+1,&3<t<4.\end{cases}$$

For $0<t<1$, both factors are negative, so $v>0$ (east). For $1<t<3$, one is negative, so $v<0$ (west). For $3<t<4$, both are positive, so $v>0$ (east). Both zero-speed instants are reversals because the velocity changes sign across each one.

(c) Evaluate the position and its derivative separately.

$$\boxed{x(2)=2\ \mathrm{m},\quad v(2)=-3\ \mathrm{m/s},\quad s(2)=3\ \mathrm{m/s}}.$$ The point is east of the origin but its coordinate is decreasing.

(d) Split the path at the actual reversals.

The positions at 0, 1, 3 and 4 s are $0,4,0,4\ \mathrm{m}$. Motion is monotone between successive listed times, so $$D=4+4+4=\boxed{12\ \mathrm{m}},\qquad\Delta x=4-0=\boxed{+4\ \mathrm{m}}.$$ Dividing by 4 s gives $\boxed{\bar s=3\ \mathrm{m/s}}$ and $\boxed{\bar v=+1\ \mathrm{m/s}}$. Endpoint separation records only the net change. It omits the intermediate westward leg and the eastward travel needed to recover that loss of net progress.