Velocity and Acceleration — Question 9

Question 9 – Solution

Strategy. Displacement uses endpoints; distance integrates speed. These measure different features of motion.

Step 1: Displacement $$\mathbf r(0)=\left\langle 3,0,0\right\rangle,\qquad \mathbf r(2\pi)=\left\langle 3,0,8\pi\right\rangle.$$ Thus $$\boxed{\Delta\mathbf r=\left\langle 0,0,8\pi\right\rangle}.$$ The elapsed time is $2\pi$, so $$\boxed{\mathbf v_{\mathrm{avg}}=\frac{\Delta\mathbf r}{2\pi}=\left\langle 0,0,4\right\rangle}.$$

Step 2: Distance and average speed $$\mathbf v(t)=\left\langle -3\sin t,3\cos t,4\right\rangle,$$ so $\|\mathbf v(t)\|=\sqrt{9+16}=5$. Hence $$\boxed{D=\int_0^{2\pi}5\,dt=10\pi},$$ and $$\boxed{\text{average speed}=\frac{10\pi}{2\pi}=5}.$$

Step 3: Interpretation During one revolution, the horizontal coordinates return to their starting values, so their net changes cancel in displacement. Nevertheless the particle continuously moves horizontally around the circle, adding positive length to the traveled path. Therefore distance $10\pi$ exceeds displacement magnitude $8\pi$.