Average Velocity and Average Speed — Practice Problem Solutions

1. Same direction the whole trip, so average velocity $=\dfrac{150\text{ mi}}{3\text{ hr}}=50\text{ mph}$ (in that direction).

2. Average speed $=\dfrac{400\text{ m}}{80\text{ s}}=5\text{ m/s}$. Average velocity $=0$ (displacement is zero — same start and finish point).

3. Distance $=60+20=80\text{ km}$, time $=1.5\text{ hr}$, average speed $\approx53.3\text{ km/h}$. Displacement $=60-20=40\text{ km east}$, average velocity $\approx26.7\text{ km/h east}$.

4. Total distance $=10+8+12=30\text{ km}$, total time $=15+10+20=45\text{ min}=0.75\text{ hr}$. Since every leg is the same direction, average velocity $=\dfrac{30}{0.75}=40\text{ km/h}$ in that direction. (Averaging the three individual speeds directly would give the wrong answer — you must use total distance over total time.)

5. Average speed $=\dfrac{200\text{ m}}{180\text{ s}}\approx1.11\text{ m/s}$. Average velocity $=0$ (the swimmer ends back at the starting wall).

6. a) Total distance $=12+12=24\text{ km}$, total time $=100\text{ min}=1.667\text{ hr}$, average speed $=\dfrac{24}{1.667}\approx14.4\text{ km/h}$. b) Average velocity $=0$ (displacement is zero — the cyclist arrives back home). c) The reasoning is flawed: average velocity only depends on net displacement, which is zero for ANY round trip regardless of how fast either leg was ridden. A zero average velocity says nothing about the speed of the return leg — it would be zero even if the cyclist rode home twice as fast.

7. Reasoning first: the hike covers the same net displacement in more time, so average velocity (displacement/time) must decrease. Displacement $=1.5\text{ km/h}\times2\text{ hr}=3\text{ km north}$ (fixed). New average velocity $=\dfrac{3\text{ km}}{3\text{ hr}}=1.0\text{ km/h north}$ — smaller, as predicted.