Motorcycle Velocity Difference and Acceleration Problem

[Caitlin.Lolz.]

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, motorcycle A has an average acceleration of 1.6 m/s2 due east, while motorcycle B has an average acceleration of 3.6 m/s2 due east. By how much did the speeds differ at the beginning of the four-second interval? And which motorcycyle is traveling faster?

I tried subtracting 1.6 from 3.6, but it didn't accept 2 as the correct answer.
What am I doing wrong?

 

[hp-p00nst3r]

For motorcycle A, in 4 seconds, it gained an extra 6.4m/s (4)(1.6)
For motorcycle B, in 4 seconds, it gained an extra 14.4m/s (4)(3.6)
Since after the 4 second interval, the 2 motorcycles have the same velocity, I can say that
v_A + 6.4 = v_B + 14.4
The difference between the 2 bikes' velocity would be v_A - v_B
By rearranging the above equation, we get v_A - v_B = 8m/s.
Since v_A > v_B initially, this must mean v_A is traveling faster.

I think what you did wrong is that when you just did 3.6 - 1.6, it only accounted for a 1-second interval. Over 4 seconds would yield a value of 8, which is what I got.

I believe this is how to do the question. If that is wrong, I apologize for that.