Average Velocity and Instantaneous Velocity — Question 7

Question 7

Rest requires simultaneous component zeros

For $t\geq0$, let $u=t/(1\ \mathrm s)$ and $\vec r(t)=(u^2-2u,\ u^3/3-u^2)$ m.

1. Find both velocity components and every time at which either component vanishes.

2. Is the particle ever completely at rest? Find the speed at 1 s.

3. Find the average velocity from 0 to 2 s and explain why its horizontal component can vanish without an interval of rest.

Question 7 - Solution

Rest requires simultaneous component zeros

(a) Differentiate and solve each component equation.

$$v_x=(2u-2)\,\mathrm{m/s},\quad v_y=u(u-2)\,\mathrm{m/s},\quad Z_x=\{1\},\quad Z_y=\{0,2\}$$ $\vec v=(2u-2,u^2-2u)$ m/s. The horizontal root is $u=1$; vertical roots are $u=0$ and $u=2$. They are different times.

(b) Intersect, rather than combine, the root sets.

$$Z_x\cap Z_y=\varnothing,\quad\boxed{\vec v(1\,\mathrm s)=(0,-1)\,\mathrm{m/s},\quad v=1\,\mathrm{m/s}}$$ There is no common root, so the particle is never at rest for $t\geq0$. At 1 s its velocity is $(0,-1)$ m/s and its speed is 1 m/s. A vanishing component alone indicates alignment of the velocity with the other axis.

(c) Use the endpoint difference.

$$\vec v_{\mathrm{av}}=\frac{(0,-4/3)-(0,0)}2=(0,-2/3)\,\mathrm{m/s}$$ $\vec r(0)=(0,0)$ m and $\vec r(2\ \mathrm s)=(0,-4/3)$ m. Average velocity is $(0,-2/3)$ m/s. The horizontal motion reverses and cancels in displacement; it is not absent throughout the interval.