Question
 A man walks on a straight road from his home to a market $2.5$ km away with a speed of $5 km^{ h -1}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5 km h ^{-1}$. What is the Average speed of the man over the interval of time
i. $0$ to $30$ min ,
ii. $0$ to $50$ min ,
iii. $0$ to $40$ min ?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
 

Answer

i. 0 to 30 min ,
Average velocity $=$ Displacement/Time $=2.5 /(1 / 2)=5 km / h$
Average speed $=$ Distance $/$ Time $=2.5 /(1 / 2)=5 km / h$
ii. 0 to 50 min
Time $=50 min=50 / 60=5 / 6 h$
Net displacement $=0$
Total distance $=2.5+2.5=5 km$
Average velocity $=$ Displacement/Time $=0$
Average speed $=$ Distance $/$ Time $=5 /(5 / 6)=6 km / h$
iii. 0 to 40 min
Speed of the man $=7.5 km / h$
Distance travelled in first $30 min=2.5 km$
Distance travelled by the man (from market to home) in the next 10 min
$=7.5 \times 10 / 60=1.25 km$
Net displacement $=2.5-1.25=1.25 km$
Total distance travelled $=2.5+1.25=3.75 km$
Average velocity $=$ Displacement/Time $=1.25 /(40 / 60)=1.875 km / h$
Average speed $=$ Distance $/$ Time $=3.75 /(40 / 60)=5.625 km / h$

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