Question

Two soccer players start from rest, 36 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.58 m/s2. The second player’s acceleration has a magnitude of 0.42 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

Answer 1

a) Both players are moving by uniformly accelerated motion, and we can write the position at time t of each of the two players as follows:

`x_1(t)= (1)/(2)a_1 t^2`

`x_2(t)=d- (1)/(2)a_2 t^2`
where

`a_1 = 0.58 m/s^2` is the acceleration of the first player

`a_2=0.42 m/s^2` is the acceleration of the second player

`d=36 m` is the initial distance between the two players
and where I put a negative sign in front of the acceleration of the second player, since he's moving in the opposite direction of the first player.

The time t at which the two players collide is the time t at which
`x_1 = x_2`, therefore:

`(1)/(2)a_1 t^2 = d- (1)/(2)a_2 t^2`
from whic we find

`t= \sqrt{ (2d)/(a_1+a_2) }= \sqrt{ (2 \cdot 36 m)/(0.58 m/s^2+0.42 m/s^2) }=8.5 s`

b) We can use the equation of
`x_1(t)` to find how far the first player run in t=8.5 s:

`x_1(t)= (1)/(2)a_1 t^2= (1)/(2)(0.58 m/s^2)(8.5 s)^2=21.0 m`