Constant Acceleration - A Special Case — Question 8

Question 8

The smallest separation can occur before either cart stops

At $t=0$ a leading point cart is $d>0$ metres east of a following point cart. The leader continues east at $10\ \mathrm{m/s}$. The follower initially moves east at $14\ \mathrm{m/s}$ and immediately begins constant acceleration $-2\ \mathrm{m/s^2}$ until it first stops; afterward it stays stationary. Set the follower’s initial position to zero. End the model at first contact if contact occurs.

1. Derive the separation $g(t)=x_{\mathrm{leader}}-x_{\mathrm{follower}}$ during the follower’s braking stage. State that stage’s time domain.

2. For $d=20\ \mathrm{m}$, find the smallest separation, when it occurs, and both velocities then. Check the later stationary-follower stage before concluding that no contact occurs.

3. Find the condition on $d$ for strictly positive separation at all times. What occurs at the boundary value of $d$?

4. For $d=3\ \mathrm{m}$, find the first contact time. The quadratic also has another root: why is it not a second prediction of this contact-terminated model?

Question 8 - Solution

The smallest separation can occur before either cart stops

(a) Subtract positions evaluated at the same time.

The follower stops when $14-2t=0$, or at 7 s. During $0\leq t\leq7$, with $t$ numerical seconds and $d$ numerical metres, $$x_F=14t-t^2,\quad x_L=d+10t,\quad \boxed{g(t)=d-4t+t^2=(t-2)^2+d-4\ \mathrm{m}}.$$ These expressions apply only up to first contact, if it occurs sooner.

(b) Check the interior minimum and the later stage.

$$g'(t)=-4+2t=0\Rightarrow t=2\ \mathrm{s};\quad g''=2>0,\quad g(t\geq7)=d+10t-49.$$

For $d=20\ \mathrm{m}$, the completed square shows the minimum is $\boxed{16\ \mathrm{m}}$ at $\boxed{t=2\ \mathrm{s}}$, inside the braking interval. Both velocities then equal 10 m/s. Before this time the follower closes the gap; afterward its lower velocity lets the gap grow.

At 7 s the follower stops at $14(7)-7^2=49\ \mathrm{m}$. The leader is at 90 m, leaving 41 m. Later the follower stays at 49 m while the leader continues east, so separation grows. The earlier minimum is the full-history minimum.

(c) An interior contact threshold is stronger than an endpoint check.

For general $d$, the minimum during braking is $d-4$. Strictly positive separation requires $\boxed{d>4\ \mathrm{m}}$. At $d=4\ \mathrm{m}$, the points touch at 2 s with equal velocities. Zero separation is contact even though the extrapolated separation would not change sign there. For $d>4$, the later stage also only increases separation.

(d) Apply the contact termination to the two roots.

$$d=3:\quad g(t)=(t-1)(t-3)=0\Rightarrow t\in\{1,3\}\ \mathrm{s};\quad t_{\mathrm{first}}=1\ \mathrm{s}.$$

At $d=3$, $g=t^2-4t+3=(t-1)(t-3)$, giving algebraic roots 1 and 3 s. The valid first contact is $\boxed{t=1\ \mathrm{s}}$, when both positions are 13 m. Continuing to the second root would extrapolate through an interval of overlapping order without a supplied post-contact rule. The model ends at 1 s, so it makes no second-contact prediction.