Question.
Abdul while driving to school computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 40 km h–1. What is the average speed for Abdul's trip ?
Abdul while driving to school computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 40 km h–1. What is the average speed for Abdul's trip ?
Solution:
Let one way distance for his trip be S.
Let $t_{1}$ be the time for his trip from home to school and $t_{2}$ be the time for his return trip.
Then $\mathrm{t}_{1}=\frac{\mathrm{S}}{\mathrm{v}_{1}}=\frac{\mathrm{S}}{20} \mathrm{~h}$, and
$\mathrm{t}_{2}=\frac{\mathrm{S}}{\mathrm{v}_{2}}=\frac{\mathrm{S}}{40} \mathrm{~h}$
Therefore, total time of trip is
$\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{\mathrm{S}}{20}+\frac{\mathrm{S}}{40}=\frac{3 \mathrm{~S}}{40} \mathrm{~h}$
Total distance covered $=2 \mathrm{~S}$
Therefore, average speed of Abdul
$\mathrm{V}_{\mathrm{av}}=\frac{\text { total distance }}{\text { total time }}=\frac{2 \mathrm{~S} \times 40}{3 \mathrm{~S}}=26.6 \mathrm{kmh}^{-1}$
Let one way distance for his trip be S.
Let $t_{1}$ be the time for his trip from home to school and $t_{2}$ be the time for his return trip.
Then $\mathrm{t}_{1}=\frac{\mathrm{S}}{\mathrm{v}_{1}}=\frac{\mathrm{S}}{20} \mathrm{~h}$, and
$\mathrm{t}_{2}=\frac{\mathrm{S}}{\mathrm{v}_{2}}=\frac{\mathrm{S}}{40} \mathrm{~h}$
Therefore, total time of trip is
$\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{\mathrm{S}}{20}+\frac{\mathrm{S}}{40}=\frac{3 \mathrm{~S}}{40} \mathrm{~h}$
Total distance covered $=2 \mathrm{~S}$
Therefore, average speed of Abdul
$\mathrm{V}_{\mathrm{av}}=\frac{\text { total distance }}{\text { total time }}=\frac{2 \mathrm{~S} \times 40}{3 \mathrm{~S}}=26.6 \mathrm{kmh}^{-1}$