Average Velocity and Instantaneous Velocity — Question 1

Question 1

Average velocity belongs to the whole interval

A walker travels straight from O $(0,0)$ m to P $(30,0)$ m in 10 s, then straight to Q $(30,40)$ m in 20 s. There are no pauses. Average velocity is $\vec v_{\mathrm{av}}=\Delta\vec r/\Delta t$; average speed is traveled distance divided by elapsed time.

1. Find the average velocity vector, its magnitude and direction for the full trip.

2. Find the average speed and explain the difference.

3. If a pause of 10 s is inserted at P, which numerators change and what are the new averages?

Question 1 - Solution

Average velocity belongs to the whole interval

(a) Use the complete interval endpoints.

$$\vec v_{\mathrm{av}}=\frac{(30,40)}{30}=(1,4/3)\,\mathrm{m/s},\quad\boxed{|\vec v_{\mathrm{av}}|=5/3\,\mathrm{m/s}}$$ $\Delta\vec r=(30,40)$ m and $\Delta t=30$ s, so $\vec v_{\mathrm{av}}=(1,4/3)$ m/s. Its magnitude is $5/3\approx1.667$ m/s, directed $53.130^\circ$ north of east.

(b) Use route length for speed.

$$\bar v_{\mathrm{speed}}=\frac{30+40}{10+20}=\frac73\,\mathrm{m/s}>\frac{\sqrt{30^2+40^2}}{30}$$ The distance is 70 m, giving average speed $70/30=7/3\approx2.333$ m/s. It exceeds the average-velocity magnitude because the bent route is longer than the 50 m endpoint separation.

(c) Include stopped time in the denominator.

$$\vec v_{\mathrm{av}}'=\frac{(30,40)}{40}=(.75,1)\,\mathrm{m/s},\quad\bar v_{\mathrm{speed}}'=70/40=1.75\,\mathrm{m/s}$$ Neither displacement nor traveled distance changes during the pause. The total duration becomes 40 s: average velocity is $(0.75,1)$ m/s, magnitude 1.25 m/s, and average speed is 1.75 m/s. An interval average includes all time between its endpoints.