I need help with velocity.?

Question

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? Which motorcycle is traveling faster?

I'm confused on how to do this. Any help is awesome!!

Answer 1

Velocity equations for 2 motorcycles:

for A: v1(t) = 1.6 t + u1

for B: v2(t) =  3.6 t + u2

Therefore

v2 - v1 = (3.6 - 1.6)t + (u2 - u1)

u2 - u1 = (v2 - v1) - 2t

At t = 4s, v2 - v1 = 0. Thus

u2 - u1 = -2*4 = -8 m/s

At the beginning, B is slower than A by 8 m/s. But after 4 s B begins to move faster, due to greater acceleration.



  

Answer 2

V(final)= V(initial) + time x acceleration

Because the bikes end up going the same speed their final velocities are equal to each other

V(final)Bike B = V(final) Bike A

Then expand the equation

V(initial B) + 4 (seconds)  * 3.6 (m/s2) = V (initial A) + 4 (seconds) * 1.6 (m/s2)

Start to simplify and solve for V(A)

V (B) + 14.4 = V (A) + 6.4

V (B) + 8 = V (A)

As you can see Bike A was going 8 m/s faster than Bike B