Question

A train travels between two stations P and Q. It travels with a velocity of $20km{h^{ - 1}}$ from P to Q and returns to station P with a velocity $30km{h^{ - 1}}$. The average speed for the total journey.
(A) $25km{h^{ - 1}}$
(B) $24km{h^{ - 1}}$
(C) $6.67m{s^{ - 1}}$
(D) $0m{s^{ - 1}}$

Answer 1

Hint: Average velocity is the ratio of total distance travelled to the total time taken. Find out the time taken for both the journeys. Add it. And then use it to calculate average speed.

Complete step by step answer:
We know that,
$s = vt$
Where,
$s$ is distance
$v$ is speed
$t$ is time taken
Let us assume that the train is travelling for P to Q with velocity ${v_1} = 20km{h^{ - 1}}$
Let the distance between the stations P and Q be $d$
Let the time taken for the train to reach station Q from P be ${t_1}$
Then, by using the above formula, we can write
$d = {v_1}{t_1}$
$ \Rightarrow d = 20{t_1}$
Rearranging it we can write
${t_1} = \dfrac{d}{{20}}$
Now, let the train returns back to P with the velocity ${v_2} = 30km{h^{ - 1}}$
The distance between the stations would be the same. i.e. $d$
Let the train returns back to P in the time ${t_2}$
Then, using the same formula, we can write
$d = {v_2}{t_2}$
$ \Rightarrow d = 30{t_2}$
Rearranging it we can write
${t_2} = \dfrac{d}{{30}}$
Now, we know that,
Average speed, ${v_a}$ is the ratio of total distance travelled to total time taken.
The total distance travelled by the train is $d + d = 2d$
The total time taken by the train is ${t_1} + {t_2}$
$\therefore {v_a} = \dfrac{{2d}}{{{t_1} + {t_2}}}$
\[ = \dfrac{{2d}}{{\dfrac{d}{{20}} + \dfrac{d}{{30}}}}\]
By cancelling the common terms, we will get
${v_a} = \dfrac{2}{{\dfrac{1}{{10}}\left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)}}$
By rearranging and cross multiplying, we get
$ = \dfrac{{2 \times 10}}{{\dfrac{{2 + 3}}{6}}}$
$ = \dfrac{{20 \times 6}}{5}$
$ \Rightarrow {v_a} = 24km{h^{ - 1}}$
Thus, the average speed of the train will be $24km{h^{ - 1}}$

Therefore, from the above explanation, the correct answer is, option (B) $24km{h^{ - 1}}$

Note:Very important concept to understand in this question is, if the speed in different intervals is different. Then the average speed is not equal to the mean of the given speeds