Average Acceleration and Instantaneous Acceleration — Question 10

Question 10

A zero average can hide substantial acceleration

A particle has $\vec v(t)=[4\cos(\omega t),2\sin(\omega t)]$ m/s with $\omega=\pi\ \mathrm{s^{-1}}$.

1. Find average acceleration from 0 to 2 s and instantaneous acceleration at zero.

2. Find the speed at 0 and 0.5 s. Is this a constant-speed motion?

3. Is acceleration ever the zero vector? Explain why its separate component zeros do not suffice.

Question 10 - Solution

A zero average can hide substantial acceleration

(a) Distinguish a periodic endpoint difference from a derivative.

$$\vec a_{\mathrm{av}}=\frac{\vec v(2)-\vec v(0)}2=\vec0,\quad\boxed{\vec a(0)=(0,2\pi)\,\mathrm{m/s^2}}$$ $\vec v(0)=\vec v(2\ \mathrm s)=(4,0)$ m/s, so average acceleration over that interval is zero. Differentiation gives $\vec a=(-4\pi\sin\omega t,2\pi\cos\omega t)$ m/s$^2$, hence $\vec a(0)=(0,2\pi)$ m/s$^2$.

(b) Check the unequal velocity amplitudes.

$$v^2=16\cos^2\omega t+4\sin^2\omega t:\quad v(0)=4,\quad v(.5)=2\quad\mathrm{m/s}$$ The two speeds are 4 m/s and 2 m/s. In general $v^2=16\cos^2\omega t+4\sin^2\omega t$, which is not constant. A periodic velocity is not automatically uniform-speed motion.

(c) Require simultaneous sine and cosine zeros.

$$a^2=16\pi^2\sin^2\omega t+4\pi^2\cos^2\omega t\ge4\pi^2\,\mathrm{m^2/s^4}>0$$ Zero acceleration would require both $\sin\omega t=0$ and $\cos\omega t=0$, which never happens. Its direction and magnitude vary, but it is nonzero at every time. The zero full-period average represents cancellation of vector changes across the interval.