Help with an acceleration physics question!?

Question

Two football players start 37m apart. One football player goes from rest (0m/s) and accelerates at 5m/s^2 towards the other player. The other player maintains a constant velocity of 3.1m/s towards the first player.

How long will it take for both players to meet each other and how far will have each player travelled?

Please explain in detail and show the steps.

Answer 1

The key is to calculate the time to impact (t), which will be the same for both players.

For player one d = 1/2 at^2; where a = 5 m/sec^2 and t is the time to impact. For player two D = vt; where v = 3.1 m/sec. Let S = 37 m, the starting distance between the players. Then S = D + d = 37, which says the sum of the two distances covered by the players in time t will close the separation S.

t = sqrt(2d/a) = v/D = t = v/(S - d); so that, sqrt(2d/a) = v/(S - d) and 2d/a = v^2/(S - d)^2. Then (S^2 -2Sd + d^2)(2d/a) = v^2; where S = 37 m, a = 5 m/sec^2, and v = 3.1 m/sec, so you can solve for d. It's messy, but solvable. Once you have d, you can solve for D from S = D + d.

The time to impact is then trivial. t = v/D will do it.