To find the displacement of the object over the given time intervals, we need to integrate the velocity function \( v(t) = -32t + 180 \) with respect to time \( t \).
a. Displacement on [2, 6]:
To find the displacement over the interval [2, 6], we integrate the velocity function:
\[ \int_{2}^{6} (-32t + 180) dt \]
Integrating with respect to time gives:
\[ \left[ -16t^2 + 180t \right]_{2}^{6} \]
\[ = [-16(6)^2 + 180(6)] - [-16(2)^2 + 180(2)] \]
\[ = [-576 + 1080] - [-64 + 360] \]
\[ = 504 - 296 \]
\[ = 208 \text{ ft} \]
Therefore, the displacement of the object over the interval [2, 6] is 208 ft.
b. Displacement on [0, 25]:
To find the displacement over the interval [0, 25], we integrate the velocity function:
\[ \int_{0}^{25} (-32t + 180) dt \]
Integrating with respect to time gives:
\[ \left[ -16t^2 + 180t \right]_{0}^{25} \]
\[ = [-16(25)^2 + 180(25)] - [-16(0)^2 + 180(0)] \]
\[ = [-10000 + 4500] - [0] \]
\[ = -5500 \text{ ft} \]
Therefore, the displacement of the object over the interval [0, 25] is -5500 ft.
