Relative Motion in One Dimension — Question 8

Question 8

Relative acceleration can first widen and then close a gap

For t in seconds, a pursuer has $x_P=2t+t^2$ m and a target has $x_T=18+5t$ m for $t\geq0$. Both remain on the same line.

1. Find the relative position, velocity and acceleration of P with respect to T.

2. Find the encounter time and both velocities then.

3. Does the separation decrease immediately? Find its largest value before the encounter.

Question 8 - Solution

Relative acceleration can first widen and then close a gap

(a) Subtract corresponding kinematic quantities.

$$x_{P/T}=t^2-3t-18,\quad v_{P/T}=2t-3,\quad\boxed{a_{P/T}=2\,\mathrm{m/s^2}}$$ $x_{P/T}=t^2-3t-18$ m, $v_{P/T}=2t-3$ m/s and $a_{P/T}=2$ m/s$^2$. The initial relative velocity is negative, so the pursuer initially loses ground despite accelerating.

(b) Solve the relative-position equation.

$$(t-6)(t+3)=0\ \Rightarrow\ t=6\,\mathrm s,\quad x=48\,\mathrm m,\quad v_P=14,\ v_T=5\quad\mathrm{m/s}$$ $t^2-3t-18=(t-6)(t+3)=0$. The physical encounter is t=6 s at x=48 m. Pursuer velocity is 14 m/s; target velocity is 5 m/s, giving relative velocity 9 m/s at the crossing.

(c) Find the turn in the gap trend.

$$d=18+3t-t^2,\quad\dot d=3-2t=0\ \Rightarrow\ t=1.5\,\mathrm s,\quad d_{\max}=20.25\,\mathrm m$$ The positive separation before crossing is $x_T-x_P=18+3t-t^2$. It grows until its derivative $3-2t$ vanishes at 1.5 s, then shrinks. The maximum gap is $18+4.5-2.25=20.25$ m. Acceleration toward catching does not require immediate closing.