Problem 5. (20 points) A particle moves on a straight line with velocity $9-t^{2} \mathrm{~cm} / \mathrm{sec}$. Find the displacement and the total distance traveled by the particle over the time interval $0 \leq t \leq 4$. Please express your final answer as a single number (not a sum or difference of numbers).

Step 1 ：First, we need to find the displacement of the particle. The displacement is the integral of the velocity function over the given time interval. So, we have \(\int_{0}^{4} (9 - t^2) dt\).

Step 2 ：Using the power rule for integration, we get \(\int_{0}^{4} (9 - t^2) dt = [9t - \frac{1}{3}t^3]_{0}^{4}\).

Step 3 ：Substituting the limits of integration, we get \([9(4) - \frac{1}{3}(4)^3] - [9(0) - \frac{1}{3}(0)^3] = 36 - \frac{64}{3} = \frac{44}{3}\) cm.

Step 4 ：For the total distance traveled, we need to consider the absolute value of the velocity because the particle can move in the opposite direction. The velocity changes sign at \(t = \sqrt{9} = 3\) seconds.

Step 5 ：So, the total distance traveled is \(\int_{0}^{3} (9 - t^2) dt + \int_{3}^{4} (t^2 - 9) dt\).

Step 6 ：Calculating the integrals, we get \([9t - \frac{1}{3}t^3]_{0}^{3} + [\frac{1}{3}t^3 - 9t]_{3}^{4} = (27 - 9) + (\frac{64}{3} - 36) = 18 + \frac{28}{3} = \frac{82}{3}\) cm.

Step 7 ：Thus, the displacement of the particle over the time interval \(0 \leq t \leq 4\) is \(\boxed{\frac{44}{3}}\) cm and the total distance traveled by the particle over the same time interval is \(\boxed{\frac{82}{3}}\) cm.