Question

As a physicist hurries to a bus stop, a bus passes her, stops ahead, and begins loading passengers. She begins running at 6 m/s. to catch the bus; however, when she is 8 meters behind the bus's sisde door, it closes and the bus begins driving away with a constant acceleration of 2 m/s^2. Relunctant to give up, she keeps running at 6 m/s until finally she reaches the bus's side door. How much time does it take for her to catch up?

Answer 1

Let after the bus moves, she is able to catch the bus after the bus travels a distance x. Now, she has to travel distance x +8 in the same time as the bus travels a distance x.

Her speed, s = 6 m/s, distance = x+8. Therefore, time taken, t = (x+8)/6 --------------- eq 1

Here for the bus, initial velocity, u = 0, a = 2m/s², distance = x

From second equation of motion, we have, x = ut+(1/2)at²

=> x = t²

Putting the value of x in equation 1, we have, t = (t²+8)/6

solving for t, we get, t = 2 or t = 4 sec.