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:
$x_1(t)= \frac{1}{2}a_1 t^2$
$x_2(t)=d- \frac{1}{2}a_2 t^2$
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:
$\frac{1}{2}a_1 t^2 = d- \frac{1}{2}a_2 t^2$
from whic we find
$t= \sqrt{ \frac{2d}{a_1+a_2} }= \sqrt{ \frac{2 \cdot 36 m}{0.58 m/s^2+0.42 m/s^2} }=8.5 s$

b) We can use the equation of $x_1(t)$ to find how far the first player run in t=8.5 s:
$x_1(t)= \frac{1}{2}a_1 t^2= \frac{1}{2}(0.58 m/s^2)(8.5 s)^2=21.0 m$