Two motorcycles are traveling due east with different velocities. However, 3.63 seconds later, they have the same velocity. During this 3.63-second interval, motorcycle A has an average acceleration of 4.55 m/s2 due east, while motorcycle B has an average acceleration of 18.9 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 3.63-second interval, and (b) which motorcycle was moving faster

52.095 m/s

Motorcycle a was moving faster

Explanation:

We start by using one of the equations of motion

V = u + at

If the first motorcycle starts with an initial speed of u(a) and accelerates at a value of a(a) = 4.55 m/s², then the final speed after a time of 3.63 seconds is V(a). We then represent it as

V(a) = u(a) + a(a).t

If the second motorcycle starts with an initial speed of u(b) and accelerates at a value of a(b) = 18.9 m/s², then the final speed after a time of 3.63 seconds is V(b). We then represent it as

V(b) = u(b) + a(b).t

Assuming that the final speeds v(a) = v(b), and then subtract the equation of the second motorcycle from that of the first, we have

0 = u(a) - u(b) + a(a).t - a(b).t

-u(a) + u(b) = a(a).t - a(b).t, on rearranging, we have

u(b) - u(a) = [a(a) - a(b)]t

Since we have the values for acceleration and the time, we substitute so that

u(b) - u(a) = (4.55 - 18.9)3.63

u(b) - u(a) = -14.35 * 3.63

u(b) - u(a) = -52.095, or we rearrange to get

u(a) - u(b) = 52.095 m/s

it would be c. predictability