Let us say that:
1 = 1st player notation
2 = 2nd player notation (the opponent)
a. First let us establish the distance travelled by the 2nd player:
d2 = 13 m/s * (t + 1.5)
d2 = 13 t + 19.5
Then the distance of the 1st player:
d1 = v0 t + 0.5 a t^2 (v0 initial velocity = 0 since he started from rest)
d1 = 0.5 * 4 m/s^2 * t^2
d1 = 2 t^2
The two distances must be equal, d1 = d2:
2 t^2 = 13 t + 19.5
t^2 – 6.5 t = 9.75
Completing the square:
(t – 3.25)^2 = 9.75 + (- 3.25)^2
t – 3.25 = ±4.5
t = -1.25, 7.75
Since time cannot be negative, therefore:
t = 7.75 seconds
So he catches his opponent after 7.75 seconds.
b. Using the equation:
d1 = 2 t^2
d1 = 2 * (7.75)^2
d1 = 120.125 m
So he travelled about 120.125 meters when he catches up to his opponent.
