Average Acceleration and Instantaneous Acceleration — Question 4

Question 4

Use a dot product to decide whether speed increases

At one instant a particle has $\vec v=(3,4)$ m/s and $\vec a=(2,-1)$ m/s$^2$. For nonzero speed, differentiating $v^2=\vec v\cdot\vec v$ gives $dv/dt=(\vec v\cdot\vec a)/v$.

1. Find the speed and its instantaneous rate of change.

2. Find the acceleration component parallel to velocity and the perpendicular remainder.

3. Would reversing a at this instant reverse velocity immediately? What would it do to the speed-change rate?

Question 4 - Solution

Use a dot product to decide whether speed increases

(a) Project acceleration along the motion.

$$v=\sqrt{3^2+4^2}=5\,\mathrm{m/s},\quad\boxed{\dot v=\frac{3(2)+4(-1)}5=0.4\,\mathrm{m/s^2}}$$ $v=5$ m/s and $\vec v\cdot\vec a=6-4=2$ m$^2$/s$^3$. Thus $dv/dt=2/5=0.4$ m/s$^2$: the particle is speeding up even though the y component of acceleration is negative.

(b) Separate the two effects.

$$\vec a_\parallel=\frac{\vec v\cdot\vec a}{v^2}\vec v=\frac2{25}(3,4)=(.24,.32),\quad\vec a_\perp=(1.76,-1.32)\quad\mathrm{m/s^2}$$ The unit velocity direction is $(3/5,4/5)$. Thus $\vec a_\parallel=(0.24,0.32)$ m/s$^2$ and $\vec a_\perp=(1.76,-1.32)$ m/s$^2$. Their sum is the original a, and the remainder has zero dot product with v.

(c) Acceleration controls change, not instantaneous replacement.

$$\dot v_{\mathrm{reversed}}=\frac{\vec v\cdot(-\vec a)}v=-0.4\,\mathrm{m/s^2},\quad\lim_{\Delta t\to0}\Delta\vec v=\vec0$$ Reversing a changes $dv/dt$ to $-0.4$ m/s$^2$ while the stated instantaneous velocity remains $(3,4)$ m/s. It changes the subsequent trend and turning sense. A finite acceleration does not cause an immediate finite jump in velocity.