Question

A $10\text{hr}$ tour is made at an average speed of $40\text{km/h}$. If during the first half of the distance, the average speed of the bus was $30\text{km/h}$. What was the average speed (in $\text{km/h}$) for the second half of the trip?

[Give nearest integer value]

Answer 1

Given,

Average speed, $V_{avg}=40\text{km/h}$

Total duration of tour, $t=10\text{hr}$

So, the total distance covered,

$d=V_{avg}\times t=40\times 10=400\text{km}$

For first half bus speed was $30\text{km/h}$,

Thus time taken to cover $(\frac{d}{2}=200\text{km})$ is
$t_1=\frac{200}{30}=6.66\text{hr}$.

Thus bus covered remining $200\text{km}$ in
$t_2=10-6.66=3.34\text{hr}$

Its average speed during second half is

$V_n=\frac{200}{3.34}\approx 60\text{km/h}$

Hence, option (c) is the correct answer.

Alternate solution:

Average speed $V_{avg}=\frac{2V_1{V}_2}{V_1+V_2}$

$\Rightarrow 40=\frac{2\times 30\times V_2}{30+V_2}$

$\Rightarrow 1200+40V_2=60V_2$

$\Rightarrow V_2=60\text{km/h}$

Accepted answer : $60$