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
[tex]\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}[/tex]
Hence, the displacement is 9 ft.
