Damn, didn't see this till now. The question is as follows:
When a bus starts moving from a stationary position with a 5 m/sec2 acceleration, a passenger is 10 meters behind it. He starts running at a constant speed in an effort to catch the bus. What is the minimum speed (in meters/sec) that he needs?
A simple picture explaining what's going on
P---------B--------------Meet
0 m----10 m------------x m
Let the passenger starts at 0 m and the bus starts at 10 m mark
Let t be the time and x be the point where they meet
At time t, the bus displacement is: delta x = x - 10 = vob*t + 1/2*a*t^2 = 0 + 1/2(5)t^2 = 2.5t^2 (1) (the bus starts from rest, so vob = 0)
The velocity of the bus is vb = vob + at = 5t
The passenger displacement is: delta x = x - 0 = vop*t (2) (he runs at constant speed, so a = 0)
Substitute (2) into (1): vop*t - 10 = 2.5t^2 (3)
Now, we are looking for the minimum speed of the passenger - that is whatever speed the bus is going at time t, the passenger must also achieve that same speed in order to catch the bus. So vop = vb = 5t (4)
Substitute (4) into (3): 5t^2 - 10 = 2.5t^2
Solving for t = 2 s
Substitute t = 2 into (4) to find vop which is
10 m/s. This is the minimum speed.
