An object is moving with velocity (in $\mathrm{ft} / \mathrm{sec}) v(t)=t^{2}+1 t-12$.
Find the displacement and total distance travelled from $t=0$ to $t=7$
Displacement: $\frac{32.9}{6}$
$\mathrm{ft}$
Total Distance Travelled:

Step 1 ：The velocity of the object is given by the function \(v(t) = t^{2} + t - 12\) in ft/sec.

Step 2 ：The displacement of an object is the integral of the velocity function from the initial time to the final time. In this case, we need to integrate the function \(v(t) = t^{2} + t - 12\) from \(t = 0\) to \(t = 7\).

Step 3 ：The total distance travelled by an object is the absolute value of the integral of the velocity function from the initial time to the final time. However, since the velocity function can be negative (which would indicate the object moving in the opposite direction), we need to find the times when the velocity is zero (i.e., the roots of the velocity function), and then integrate the absolute value of the velocity function over each interval between the roots and the initial and final times.

Step 4 ：The roots of the velocity function are \(-4\) and \(3\). Therefore, the intervals of integration for the total distance travelled are \([0, 3]\) and \([3, 7]\).

Step 5 ：The displacement of the object from \(t = 0\) to \(t = 7\) is \(\frac{329}{6}\) ft, which simplifies to \(54.83\) ft.

Step 6 ：The total distance travelled by the object from \(t = 0\) to \(t = 7\) is the sum of the definite integrals of the absolute value of the velocity function over the intervals \([0, 3]\) and \([3, 7]\), which is \(99.83\) ft.

Step 7 ：Final Answer: The displacement of the object from \(t = 0\) to \(t = 7\) is \(\boxed{54.83 \, \text{ft}}\) and the total distance travelled by the object over the same period is \(\boxed{99.83 \, \text{ft}}\).