Two buses

Problem

Two buses. A bus is stopped at a traffic light; it sets off at the instant an identical bus whizzes past it at constant velocity $v_0$. The first bus accelerates uniformly with $a > 0$ while the second carries on unchanged. Between the two, which is ahead right after departure? Which arrives first at a very large distance? Is there an instant at which the two are side by side again? Discuss qualitatively, without calculations.

Solution

It helps to reason about the two equations of motion, with the origin at the traffic light and time starting from the instant of the overtaking: the accelerating bus advances according to $x_1 = \tfrac{1}{2} a t^2$, the one at constant velocity according to $x_2 = v_0 t$.

Right after departure. For small times the term $\tfrac{1}{2} a t^2$ (which grows as the square of time) is negligible compared with $v_0 t$ (which grows linearly). Indeed, the first bus starts from zero velocity and needs time to pick up speed. So right away the second bus is ahead, the one that was already travelling at $v_0$.

At a very large distance. As time passes, the quadratic term prevails over the linear one: the accelerating bus keeps increasing its own velocity and inevitably ends up overtaking the other. In the long run the first bus arrives first, the one that accelerates.

Instant of being side by side. Since at the start the second is ahead and later the first is ahead, by continuity there must exist an instant at which the two are side by side again. Setting $x_1 = x_2$, i.e. $\tfrac{1}{2} a t^2 = v_0 t$, gives (besides $t = 0$) the instant:

$t = \frac{2 v_0}{a}$

at which the first bus catches up with the second. From that moment on it stays permanently in the lead.

$\ev{\text{initially the 2nd is ahead; in the long run the 1st; side by side at } t = 2v_0/a}$

Links

Topics: Kinematics Concepts: Uniformly accelerated motion