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

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