The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.
Writing the acceleration as a function of time:
a(t) = 1 + 3√t
Integrating acceleration, we obtain velocity:
v(t) = t + 2(t)^(3/2) + c;
object at rest so velocity at t = 0 is 0 so c = 0.
v(t) = t + 2(t)^(3/2)
Integrating velocity to obtain an equation for displacement:
d(t) = t²/2 + 4/5 t^(5/2) + c
Applying limits from t = 0 to t = 9
d = 9²/2 + 4/5 9^(5/2)
d = 234.9 m
