hi,
i tried to solve the problem, but i think that the numbers given are a little incorrect. anyways, here is my procedure:
1)Let us take the two footballers as A and B respectively, so, A is at rest and starts accelerating at 0.5m/s^2, and B moves with an uniform velocity of 3.1m/s.
2)at some point of time the two are going to crash, so take that point to be O. the distance from the starting point of A to O is x meters (let us consider). hence, the distance from the starting point of B to O will be 37 - x meters. sorry for the rough diagram.
A ---- x meters ----- O ------------ 37-x meters ----------------B
3)now, use the 2nd equation of motion to solve, s = ut + 1/2 * at^2, where u is initial velocity, and s is the distance. here the time t will remain constant because both will crash at the same time.
for A,
s = ut + 1/2 * at^2
=> x = 1/2 * 0.5t^2 ----------- (1)
for B,
s = ut + 1/2 * at^2
=> 37 - x = 3.1t ------------- (2)
you can now solve (1) and (2) to get answers to a) and c) by solving simultaneous equations for the values for t and x respectively.
4)after you get all the previous values, getting an answer for b) becomes very easy with numerous options for equations. (eg: you could use v = u + at, where v is final velocity)
