Graphical Integration in Motion Analysis — Question 4

Question 4

Equal velocities mark an extremum of separation, not a meeting

Two ideal point markers A and B move on one east-positive line and may pass through one another without interaction. At $t=0$, B is 3 m east of A. Over 0–4 s their relative velocity $w=v_B-v_A$ increases linearly from $-4$ to $+4\ \mathrm{m/s}$. Define signed separation $g=x_B-x_A$.

1. Write $g(t)$ by accumulating the relative-velocity area, and find every meeting time in the interval.

2. At what time do the velocities agree? Find $g$ then, and explain why this event is not a meeting.

3. State which marker is east of the other on each interval separated by the meetings. Find the greatest ordinary distance $|g|$ between them during the full run.

4. Replace the initial separation by a positive value $d$ while leaving the graph fixed. Classify the number of meetings for $0<d<4\ \mathrm{m}$, $d=4\ \mathrm{m}$ and $d>4\ \mathrm{m}$. Interpret the limiting meeting at $d=4\ \mathrm{m}$.

Question 4 - Solution

Equal velocities mark an extremum of separation, not a meeting

(a) Meeting requires the accumulated area to cancel the initial gap.

Using numerical SI values, $w=-4+2t$ and $$g(t)=3+\int_0^t(-4+2u)\,du=3-4t+t^2=(t-1)(t-3).$$ Thus $\boxed{t=1\ \mathrm{s}\text{ and }3\ \mathrm{s}}$ are the two meetings. The negative relative-velocity area closes the initial positive gap and can carry B to the west of A.

(b) Zero relative velocity stops the change of signed separation.

The graph crosses zero at $\boxed{t=2\ \mathrm{s}}$. At that instant $\boxed{g=-1\ \mathrm{m}}$: B is 1 m west of A. Signed separation is at its minimum because $w$ changes from negative to positive. Equality of velocities alone is not equality of positions.

(c) Distinguish signed separation from its magnitude.

B is east of A for $0\leq t<1$ s, west for $1<t<3$ s, and east again for $3<t\leq4$ s. They coincide at the boundary times 1 and 3 s. The signed gap ranges from $-1$ to $+3\ \mathrm{m}$, so $\boxed{\max|g|=3\ \mathrm{m}}$, attained at 0 and 4 s. The largest westward separation is only 1 m.

(d) Compare the initial gap with the largest available closing area.

$$g_d(t)=0\iff(t-2)^2=4-d;\qquad N_{\mathrm{meetings}}=\begin{cases}2,&0<d<4,\\1,&d=4,\\0,&d>4.\end{cases}$$

The negative triangle during 0–2 s has area $-4\ \mathrm{m}$. With $d$ numerically in metres, $$g_d=d-4+(t-2)^2.$$ For $0<d<4$, there are two roots $\boxed{t=2\pm\sqrt{4-d}}$ in seconds, both inside the window. For $d=4\ \mathrm{m}$ there is one meeting at 2 s; the gap touches zero without changing sign and the velocities agree there. For $d>4\ \mathrm{m}$ the gap stays positive, so there is no meeting. This is an ideal trajectory comparison, not a collision model.