The function $v(t)=9 \cos 3 t, 0 \leq t \leq 2 \pi$, is the velocity in $\mathrm{m} / \mathrm{sec}$ of a particle moving along the $x$-axis. Complete parts (a) through (c).
a. Graph the velocity function over the given interval. Then determine when the motion is in the positive direction and when it is in the negative direction.
b. Find the displacement over the given interval.
c. Find the distance traveled over the given interval.
Answer
Calculating the absolute value of the integral, we get $|3 \sin 3t \Big|_0^{2\pi}| = |3(\sin 6\pi - \sin 0)| = 0$. So, the distance traveled over the given interval is also \(\boxed{0}\) meters.
Step 1 ：First, we need to graph the velocity function $v(t)=9 \cos 3 t$ over the interval $0 \leq t \leq 2 \pi$.
Step 2 ：From the graph, we can see that the motion is in the positive direction when the velocity is positive, and in the negative direction when the velocity is negative. Therefore, we need to find the values of $t$ for which $v(t) = 0$.
Step 3 ：Solving the equation $9 \cos 3 t = 0$, we get $t = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$.
Step 4 ：So, the motion is in the positive direction when $0 \leq t < \frac{\pi}{6}, \frac{\pi}{2} < t < \frac{5\pi}{6}, \frac{7\pi}{6} < t < \frac{3\pi}{2}, \frac{11\pi}{6} < t \leq 2\pi$, and in the negative direction when $\frac{\pi}{6} \leq t < \frac{\pi}{2}, \frac{5\pi}{6} \leq t < \frac{7\pi}{6}, \frac{3\pi}{2} \leq t < \frac{11\pi}{6}$.
Step 5 ：Next, we need to find the displacement over the given interval. The displacement is the integral of the velocity function over the interval, which is $\int_{0}^{2\pi} v(t) dt = \int_{0}^{2\pi} 9 \cos 3t dt$.
Step 6 ：Calculating the integral, we get $3 \sin 3t \Big|_0^{2\pi} = 3(\sin 6\pi - \sin 0) = 0$. So, the displacement over the given interval is \(\boxed{0}\) meters.
Step 7 ：Finally, we need to find the distance traveled over the given interval. The distance traveled is the absolute value of the integral of the velocity function over the interval, which is $|\int_{0}^{2\pi} v(t) dt| = |\int_{0}^{2\pi} 9 \cos 3t dt|$.
Step 8 ：Calculating the absolute value of the integral, we get $|3 \sin 3t \Big|_0^{2\pi}| = |3(\sin 6\pi - \sin 0)| = 0$. So, the distance traveled over the given interval is also \(\boxed{0}\) meters.
