Question

Runner 1 is standing still on a straight running track. Runner 2 passes him, running with a constant speed of 1.5 m/s. Just as runner 2 passes, runner 1 accelerates with a constant acceleration of 0.73 m/s^2. How far down the track does runner 1 catch up with runner 2?

Answer 1

first, find how long it takes:
1.5t = 1/2 * 0.73 t^2
Then find the distance covered in that time.

Answer 2

1st runner: Vo = 0, a = 0.73 m/s^2.
2nd runner: Vo = 1.5 m/s. , a = 0.

When the 1st runner catches up:
d1 = d2.
Vo*t + 0,5a*t^2 = Vo*t + 0.5a*t^2.
0 + 0.5*0.73*t^2 = 1.5*t + 0,
0.365t^2 - 1.5t = 0,
t^2 - 4.11t = 0,
t(t-4.11) = 0,
t-4.11 = 0,
t = 4.11 s.