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

Step-by-step 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 [ -(t^3)/(3) + 6t \right ]_(0)^(3) \\ \\ = \left ( -((3)^3)/(3) + 6(3) \right ) - \left ( -((0)^3)/(3) + 6(0) \right ) \\ \\ \boxed{\text{displacement} = 9}`

Hence, the displacement is 9 ft.