Question

Two motorcycles are traveling due east with different velocities. However, 2.16 seconds later, they have the same velocity. During this 2.16-second interval, motorcycle A has an average acceleration of 1.30 m/s2 due east, while motorcycle B has an average acceleration of 19.4 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 2.16-second interval, and (b) which motorcycle was moving faster?

Answer 1

1.30 m/s^2 = (Vt - Vo) / 2.16 s,
Cross multiply:
Vt - Vo = 2.81 m/s,
Delta V = 2.81 m/s for Motorcycle A.

19.4 m/s^2 = (Vt - Vo) / 2.16 s,
Vt - Vo = 41.9 m/s ,
Delta V = 41.9 m/s for Motorcycle B.

41.9 m/s - 2.81 m/s = 39.1 m/s =
Difference in speed at beginning of
2.16 s interval.