Two soccer players start from rest, 33 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 0.50 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?
1/2 (.58) t^2 + 1/2 (.50) t^2 = 33
t = 7.8 sec
1st player runs 17.72 m
To find the time it takes for the players to collide, we can use the kinematic equation:
s = ut + 0.5at^2
where s is the distance between the players, u is the initial velocity (which is 0 since they start from rest), a is the acceleration, and t is the time.
For the first player:
s = 33m (distance between the players)
a = 0.58 m/s^2
Plugging in the values, we have:
33 = 0.5 * 0.58 * t^2
Simplifying the equation gives us:
0.578t^2 = 33
Now, let's solve for t:
t^2 = 33 / 0.578
t^2 = 57.09
t ≈ √57.09
t ≈ 7.56 seconds
So, it takes approximately 7.56 seconds for the players to collide.
To find how far the first player has run at the instant of collision, we can use the kinematic equation again. This time, we will find the distance traveled (s) by the first player.
s = ut + 0.5at^2
For the first player:
a = 0.58 m/s^2 (acceleration)
t = 7.56 seconds (time)
Plugging in the values, we have:
s = 0 * 7.56 + 0.5 * 0.58 * (7.56)^2
Calculating the expression gives us:
s ≈ 0 + 0.5 * 0.58 * 57.09
s ≈ 16.68 m
So, at the instant of collision, the first player has run approximately 16.68 meters.