Answer to Question #244644 in Physics for Dontworry

Question

Two soccer players start from rest, 49 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.40 m/s2. The second player’s acceleration has a magnitude of 0.50 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) Let $L=49\ m$ is the distance between two soccer players, $d$ is the distance that the first player runs before the players collide. Then, we can write the displacements of each soccer player:

$$d_1=\dfrac{1}{2}a_1t^2=d,$$ $$d_2=\dfrac{1}{2}a_2t^2=L-d.$$

Let's substitute $d$ from the first equation into the second one and find the time that passes before the players collide:

$$\dfrac{1}{2}a_2t^2=L-\dfrac{1}{2}a_1t^2,$$ $$\dfrac{1}{2}t^2(a_1+a_2)=L,$$ $$t=\sqrt{\dfrac{2L}{(a_1+a_2)}}=\sqrt{\dfrac{2\cdot49\ m}{(0.4\ \dfrac{m}{s^2}+0.5\ \dfrac{m}{s^2})}}=10.4\ s.$$

(b) We can find the distance that the first player runs before the players collide as follows:

$$d=\dfrac{1}{2}a_1t^2=\dfrac{1}{2}\cdot0.4\ \dfrac{m}{s^2}\cdot(10.4\ s)^2=21.6\ m.$$