Question

A man walks on a straight road from his home to a market $2.5km$ away with a speed of $5km{h}^{-1}$ . Finding the market closed, he instantly turns and walks back home with a speed of $7.5km{h}^{-1}$. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time
(i) $0$ to $30min$,
(ii) $0$ to $50min$
(iii) $0$ to $40min$?

Answer 1

Distance to market $s=2.5km=2.5\times {10}^3=2500m$

Speed with which he goes to market $=5km/h=5\frac{{10}^3}{3600}=\frac{25}{18}m/s$

Speed with which he comes back $=7.5km/h=7.5\times \frac{{10}^3}{3600}=\frac{75}{36}m/s$

(a)Average velocity is zero since his displacement is zero.

(b)

(i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market:$\frac{2.5}{5}=1/2h=30$ minutes.

Average speed over this interval $=5km/h$

(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :$7.5\times \frac{1}{3}=2.5km$

His average speed in 0 to 50 minutes: $V_{avg}=\frac{distancetraveled}{time}$

$=\frac{2.5+2.5}{\left(50/60\right)}=6km/h$

(iii)In 40-30=10 minutes he travels a distance of :$7.5\times \frac{1}{6}=1.25km$

$V_{avg}=\frac{2.5+1.25}{\left(40/60\right)}=5.625km/h$