Average Velocity and Instantaneous Velocity — Question 4

Question 4

A turning coordinate is not necessarily a stopped particle

With $u=t/(1\ \mathrm s)$, a particle follows $\vec r=[(u-2)^2\hat\imath+3u\hat\jmath]$ m. Consider times near $t=2$ s.

1. Find the instantaneous velocity at 2 s and decide whether the particle is at rest.

2. Describe the signs of the horizontal velocity just before and after 2 s.

3. Find the average velocity from 1 s to 3 s and compare its direction with the tangent at 2 s.

Question 4 - Solution

A turning coordinate is not necessarily a stopped particle

(a) Both velocity components must vanish for rest.

$$\vec v=(2(u-2),3)\,\mathrm{m/s}\ \Rightarrow\ \boxed{\vec v(2\,\mathrm s)=(0,3)\,\mathrm{m/s}}$$ Differentiation gives $\vec v=[2(u-2)\hat\imath+3\hat\jmath]$ m/s. At 2 s it is $3\hat\jmath$ m/s. The particle moves upward at 3 m/s even though its x coordinate has a minimum.

(b) Track reversal in one coordinate.

$$\operatorname{sgn}(v_x)=\operatorname{sgn}(u-2),\quad v_y=3\,\mathrm{m/s}>0$$ $v_x<0$ before 2 s and $v_x>0$ after it, so horizontal motion changes from leftward to rightward. Meanwhile $v_y=3$ m/s never changes sign. The full trajectory has a vertical tangent at the turning x coordinate.

(c) Evaluate the symmetric interval.

$$\vec v_{\mathrm{av}}=\frac{(1,9)-(1,3)}{2}=(0,3)\,\mathrm{m/s}=\vec v(2\,\mathrm s)$$ The two positions are $(1,3)$ and $(1,9)$ m, giving average velocity $(0,3)$ m/s. Here it equals the midpoint instantaneous velocity. Equal endpoint x coordinates mean zero average horizontal velocity, not zero horizontal velocity throughout the interval.