What is the final velocity of a motorcycle that accelerates at 3.0 ft/s^2 for 15 s from an initial velocity of 22 ft/s?

The final velocity of the motorcycle, v = 67 ft/s

Explanation:

Given that,

Acceleration of the motorcycle, a = 3.0 ft/s²

Initial velocity of the motorcycle, u = 22 ft/s

Time interval to reach final velocity, t = 15 s

The final velocity of the motorcycle, v = ?

From the equations of motion

v = u + at

Substituting the values in the above equation

v = 22 ft/s + 3.0 ft/s² · 15 s

= 67 ft/s

Hence, the final velocity of a motorcycle is v = 67 ft/s