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.

 

[baba89]

It seems like you are trying to find the difference in velocities at the beginning of the four-second interval. To do this, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval. In this case, we know that the final velocities are the same (since they have the same velocity after four seconds) and the time interval is also the same (four seconds). Therefore, we can set up the following equation:
v = u + at
where u is the initial velocity for both motorcycles.
For motorcycle A, we know that a = 1.6 m/s^2 and t = 4 seconds. Substituting these values into the equation, we get:
v = u + (1.6 m/s^2)(4 s)
v = u + 6.4 m/s
Similarly, for motorcycle B, we know that a = 3.6 m/s^2 and t = 4 seconds. Substituting these values into the equation, we get:
v = u + (3.6 m/s^2)(4 s)
v = u + 14.4 m/s
Since we know that both motorcycles have the same final velocity, we can set these two equations equal to each other:
u + 6.4 m/s = u + 14.4 m/s
Solving for u, we get u = -8 m/s.
This means that at the beginning of the four-second interval, motorcycle B was traveling 8 m/s faster than motorcycle A. Therefore, motorcycle B is traveling faster than motorcycle A.