Question 5
2 answers:
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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}](https://tex.z-dn.net/?f=%5Ctext%7Bdisplacement%7D%20%3D%20%5Cint_%7B0%7D%5E%7B3%7D%20%7Bv%28t%29%20dt%7D%0A%5C%5C%0A%5C%5C%20%3D%20%5Cint_%7B0%7D%5E%7B3%7D%20%7B%28-t%5E2%20%2B%206%29%20dt%7D%0A%5C%5C%0A%5C%5C%20%3D%20%5Cleft%20%5B%20-%5Cfrac%7Bt%5E3%7D%7B3%7D%20%2B%206t%20%5Cright%20%5D_%7B0%7D%5E%7B3%7D%0A%5C%5C%0A%5C%5C%20%3D%20%5Cleft%20%28%20-%5Cfrac%7B%283%29%5E3%7D%7B3%7D%20%2B%206%283%29%20%5Cright%20%29%20-%20%5Cleft%20%28%20-%5Cfrac%7B%280%29%5E3%7D%7B3%7D%20%2B%206%280%29%20%5Cright%20%29%0A%5C%5C%0A%5C%5C%20%5Cboxed%7B%5Ctext%7Bdisplacement%7D%20%3D%209%7D)
Hence, the displacement is 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
Hence, the displacement is 9 ft.