Average Velocity and Instantaneous Velocity — Question 9

Question 9

A half-circle average does not point along the local tangent

A particle has $\vec r(t)=[2\cos(\omega t)\hat\imath+2\sin(\omega t)\hat\jmath]$ m, with $\omega=\pi\ \mathrm{s^{-1}}$. Consider $0\leq t\leq1$ s.

1. Find the average velocity over the interval.

2. Differentiate to obtain instantaneous velocity at 0 and 1 s and the speed at every time.

3. Find the average speed over the interval and explain why it differs from the magnitude of average velocity.

Question 9 - Solution

A half-circle average does not point along the local tangent

(a) Use the diameter displacement.

$$\boxed{\vec v_{\mathrm{av}}=\frac{(-2,0)-(2,0)}1=(-4,0)\,\mathrm{m/s}}$$ The positions are $(2,0)$ m and $(-2,0)$ m. Hence $\vec v_{\mathrm{av}}=(-4,0)$ m/s, magnitude 4 m/s. It points along the chord joining the endpoints.

(b) Differentiate the trigonometric functions.

$$\vec v=2\pi(-\sin\omega t,\cos\omega t)\,\mathrm{m/s},\quad v=2\pi\sqrt{\sin^2\omega t+\cos^2\omega t}=2\pi\,\mathrm{m/s}$$ $\vec v=2\pi[-\sin(\pi t/\mathrm s)\hat\imath+\cos(\pi t/\mathrm s)\hat\jmath]$ m/s. At the two endpoints it is $(0,2\pi)$ and $(0,-2\pi)$ m/s. Its magnitude is always $2\pi$ m/s because the squared sine and cosine sum to one.

(c) Use the actual arc traveled.

$$\bar v_{\mathrm{speed}}=\frac{\pi R}{\Delta t}=2\pi\,\mathrm{m/s}>|\vec v_{\mathrm{av}}|=4\,\mathrm{m/s}$$ The particle travels half a circumference, $2\pi$ m, in 1 s, so average speed is $2\pi\approx6.283$ m/s. Constant speed does not imply constant vector velocity: the tangent direction changes throughout the half circle.