Average Acceleration and Instantaneous Acceleration — Question 5

Question 5

At rest, acceleration can still be nonzero

Let $\vec v(t)=(t-2\ \mathrm s)(2\hat\imath-\hat\jmath)\ \mathrm{m/s^2}$ near 2 s. The position is continuous and this differentiable velocity law holds on both sides.

1. Find acceleration and the instant of rest.

2. Find speed immediately before and after 2 s and its one-sided time derivatives.

3. Why must the formula $dv/dt=(\vec v\cdot\vec a)/v$ not be used directly at the rest instant?

Question 5 - Solution

At rest, acceleration can still be nonzero

(a) Differentiate the vector, not its magnitude.

$$\boxed{\vec a=\frac{d\vec v}{dt}=(2,-1)\,\mathrm{m/s^2}},\quad\vec v=\vec0\ \Leftrightarrow\ t=2\,\mathrm s$$ $\vec a=(2,-1)$ m/s$^2$ at every time, including 2 s. Both velocity components vanish at 2 s, so the particle is momentarily at rest while acceleration remains nonzero.

(b) Take the absolute value for speed.

$$v=\sqrt5\,|t-2\,\mathrm s|\,\mathrm{m/s^2},\quad\dot v_-=-\sqrt5,\quad\dot v_+=\sqrt5\quad\mathrm{m/s^2}$$ $v=\sqrt5\,|t-2\ \mathrm s|\ \mathrm{m/s^2}$. Its slope is $-\sqrt5$ m/s$^2$ before 2 s and $+\sqrt5$ m/s$^2$ afterward. Speed reaches a cusp-like minimum of zero as the velocity reverses along a fixed line.

(c) The division assumes nonzero speed.

$$\left.\frac{\vec v\cdot\vec a}{v}\right|_{t=2\,\mathrm s}=\frac00\text{ undefined},\quad\dot v_-\ne\dot v_+$$ At 2 s the ratio would be $0/0$, undefined. The velocity vector is differentiable, so acceleration exists; its magnitude need not be differentiable at a zero crossing. This example separates acceleration from the derivative of speed.