Velocity and Acceleration — Question 5

Question 5 – Solution

Strategy. Differentiate, then use the dot-product angle formula with $\mathbf k=\left\langle 0,0,1\right\rangle$.

Step 1: Motion quantities $$\boxed{\mathbf v(t)=\left\langle -6\sin 2t,6\cos 2t,4\right\rangle},$$ and $$\boxed{\mathbf a(t)=\left\langle -12\cos 2t,-12\sin 2t,0\right\rangle}.$$ The speed is $$\|\mathbf v\|=\sqrt{36(\sin^22t+\cos^22t)+16} =\boxed{2\sqrt{13}}.$$

Step 2: Direction angle If $\theta$ is the angle with the positive $z$-axis, $$\cos\theta=\frac{\mathbf v\cdot\mathbf k}{\|\mathbf v\|\|\mathbf k\|} =\frac 4{2\sqrt{13}}=\frac 2{\sqrt{13}}.$$ Therefore $$\boxed{\theta=\cos^{-1}\left(\frac 2{\sqrt{13}}\right)}.$$ No $t$ occurs, so both speed and angle are constant.

Step 3: Acceleration geometry Its horizontal component is $$-4\left\langle 3\cos 2t,3\sin 2t,0\right\rangle,$$ which is opposite the radial position from the $z$-axis. Its $z$-component is zero, so acceleration points horizontally inward. Also $\mathbf v\cdot\mathbf a=0$, consistent with constant speed.