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.