Let x = the distance that the first player moves before the crunch.
Then, the distance that the second player moves is 37-x.
The time it takes for the first player to move x metres is found out with the formula: Distance = Initial Velocity * time + 1/2 * acceleration * time²
x = 0 + 1/2 (0.5m/s²)t²
x = (0.25m/s²)t²
Now let's find it out for the second player.
Distance = Velocity * time, since it's a constant velocity.
37-x = (3.1m/s)t
solve this for x
x = 37 - (3.1m/s)t
Now we have two equations with x = something, so we can equate the two together.
(0.25m/s²)t² = 37 - (3.1m/s)t
0.25t² + 3.1t - 37 = 0
Solve this with the quadratic equation
t = 7.5s, that's the answer for A).
Now how fast was the accelerating player going?
Final velocity = Initial velocity + Acceleration * Time
Final Velocity = 0 m/s + (0.5m/s²)(7.5s) <use the number before the roundoff>
= 3.7 m/s
That's the answer for B
Now just plug in the time into x = (0.25m/s²)t²
x = (0.25m/s²)(7.5s)²
= 14m
That's how far the accelerating player ran.
The other player ran 37m - 14m = 23m.
TO RECAP
A) 7.5s
B) 3.7m/s
C) 14m, 23m.
