Average Acceleration and Instantaneous Acceleration — Question 7

Question 7

An abrupt velocity corner has no finite instantaneous acceleration

Velocity is $\vec v=(2,t/\mathrm s)$ m/s for $t<0$, and $\vec v=(2,2+t/\mathrm s)$ m/s for $t>0$. At zero the vertical velocity jumps by 2 m/s. Treat this as an idealized recorded jump, not a finite-acceleration transition.

1. Find acceleration at every nonzero time near zero.

2. Find average acceleration over $[-h,h]$ for $h>0$.

3. Does this average approach a finite acceleration at zero? Explain why equal accelerations on either side are insufficient.

Question 7 - Solution

An abrupt velocity corner has no finite instantaneous acceleration

(a) Differentiate each smooth branch.

$$\boxed{\vec a(t)=(0,1)\,\mathrm{m/s^2}\quad(t\ne0)}$$ Both branches have $a_x=0$ and $a_y=1$ m/s$^2$. These finite derivatives describe only times strictly before or after the discontinuity.

(b) Include the velocity jump in the interval difference.

$$\vec a_{\mathrm{av}}=\frac{(2,2+\delta)-(2,-\delta)}{2\delta}\,\mathrm{m/s^2}=(0,1+1/\delta)\,\mathrm{m/s^2}$$ Let $\delta=h/(1\ \mathrm s)$. Endpoint velocities are $(2,-\delta)$ and $(2,2+\delta)$ m/s. The average is $(0,1+1/\delta)$ m/s$^2$. The fixed 2 m/s jump is divided by a shrinking interval.

(c) Inspect the limiting behavior.

$$\lim_{\delta\to0^+}(1+1/\delta)=+\infty,\quad\vec v(0^+)-\vec v(0^-)=(0,2)\,\mathrm{m/s}$$ As $h\to0^+$, the vertical average grows without bound. No finite instantaneous acceleration exists at zero for this ideal history. Equal derivatives away from a jump do not make the velocity continuous or differentiable at the jump. A real finite-duration transition would require its own resolved velocity history.