Answer to Question #90626 in Classical Mechanics for Raj

Question

A pedestrian is running at his maximum speed of 6.0 m/sec to catch a bus stopped at a traffic light. When his is 15m from the bus, the light changes and the bus accelerates uniformly at 1m/s/s. Does he make it to the bus? If so , how far does he have to run in order to catch it? If not, how close does he get?

Answer 1

Let us change the frame to the frame of the pedestrian. In this one the initial velocities of the pedestrian and the bus are zero and 6.0 m/sec in the direction of the pedestrian respectively. The initial distance between them is 15 m.

The minimal value of distance corresponds to the moment when the velocity of the bus equals zero.

$$v(t) = v_0 - at =0$$ $$t = \frac{v_0}{a}$$

where v₀=6m/sec, a=1m/sec².

The distance for uniform acceleration

$$x(t) = x_0 - v_0 t + \frac{at^2}{2}$$

using the value of t obtained above, we get

$$x = x_0-\frac{v_0^2}{a} +\frac{v^2_0}{2a} = x_0 -\frac{v_0^2}{2a} = 15\, \mathrm{m} - \frac{36}{2}\, \mathrm{m} = -3\, \mathrm{m}$$

So, the pedestrian will catch the bus. Let us find the corresponding time, when x=0

$$x_0 - v_0 t + \frac{at^2}{2} =0$$ $$t = \frac{1}{a} \bigg(v_0\pm\sqrt{v_0^2 - 2x_0 a}\bigg)$$

As we know, in the frame of the pedestrian the bus goes by the pedestrian and then turns back. We are looking for the first moment, so we choose the minus sign. It gives us

$$t = \frac{1}{1} \bigg(6 - \sqrt{36 - 2\cdot 15}\bigg) \, \mathrm{m} = (6-\sqrt{6}) \, \mathrm{m}$$

In the initial frame the pedestrian moves with v₀ during this time, so the distance he has to run is

$$d = v_0 t = 6 \cdot (6-\sqrt{6})\,\mathrm{m} \approx 21.3\, \mathrm{m}$$