(i) The time taken from home to office,
(ii) For the interval 0 - 50 min
In returning journey, upto 50 min he will walk for (50 - 30) = 20 min.
Therefore distance travelled in this interval of time,
s’ = v’ × t’ = 7.5 × \(\frac{20}{60}\) km
s’ = 2.5 km
Thus, in next 20 min, he will reach his home.
∴ Travelled distance = 2.5 + 2.5 =5 km
Total time 50 min = 50/ 60 = 5/6 hr
Total displacement = 0
Hence, the average velocity = \(\frac{0}{5 / 6}\) = 0 km/hr
and average speed = \(\frac{\text { Totaldistance }}{\text { Total time }}=\frac{5}{5 / 6}\)
= 6 km/hr
(iii) For time interval 0 to 40 min
Time of return journey = 40 – 30 = 10 min
= \(\frac{10}{60} \mathrm{hr}=\frac{1}{6} \mathrm{hr}\)
The distance travelled in this time = 7.5 × 1/ 6 = 1.25 km
Total time = 40 min = \(\frac{40}{60}=\frac{2}{3}\) km
Total displacement = 2.5 - 1.25 = 1.25 km.
Hence, average speed.
