Question

For an object whose velocity in ft/sec is given by v(t) = −t^2+ 6, what is its displacement, in feet, on the interval t = 0 to t = 3 secs?

Answer 1

Answer: 9 ft

Explanation: Since the velocity is the derivative of position vector and the displacement is the difference between two position vectors, the displacement is computed as a definite integral of velocity for a time interval (from one time to another time). 

Because the time interval is t =0 second to t = 3 seconds, the displacement is calculated as

$\text{displacement} = \int_{0}^{3} {v(t) dt} \\ \\ = \int_{0}^{3} {(-t^2 + 6) dt} \\ \\ = \left [ -\frac{t^3}{3} + 6t \right ]_{0}^{3} \\ \\ = \left ( -\frac{(3)^3}{3} + 6(3) \right ) - \left ( -\frac{(0)^3}{3} + 6(0) \right ) \\ \\ \boxed{\text{displacement} = 9}$

Hence, the displacement is 9 ft.