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Question: A train travels between two stations P and Q. It travels with a velocity of \(20km{h^{ - 1}}\) from ...

A train travels between two stations P and Q. It travels with a velocity of 20kmh120km{h^{ - 1}} from P to Q and returns to station P with a velocity 30kmh130km{h^{ - 1}}. The average speed for the total journey.
(A) 25kmh125km{h^{ - 1}}
(B) 24kmh124km{h^{ - 1}}
(C) 6.67ms16.67m{s^{ - 1}}
(D) 0ms10m{s^{ - 1}}

Explanation

Solution

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=vts = vt
Where,
ss is distance
vv is speed
tt is time taken
Let us assume that the train is travelling for P to Q with velocity v1=20kmh1{v_1} = 20km{h^{ - 1}}
Let the distance between the stations P and Q be dd
Let the time taken for the train to reach station Q from P be t1{t_1}
Then, by using the above formula, we can write
d=v1t1d = {v_1}{t_1}
d=20t1\Rightarrow d = 20{t_1}
Rearranging it we can write
t1=d20{t_1} = \dfrac{d}{{20}}
Now, let the train returns back to P with the velocity v2=30kmh1{v_2} = 30km{h^{ - 1}}
The distance between the stations would be the same. i.e. dd
Let the train returns back to P in the time t2{t_2}
Then, using the same formula, we can write
d=v2t2d = {v_2}{t_2}
d=30t2\Rightarrow d = 30{t_2}
Rearranging it we can write
t2=d30{t_2} = \dfrac{d}{{30}}
Now, we know that,
Average speed, va{v_a} is the ratio of total distance travelled to total time taken.
The total distance travelled by the train is d+d=2dd + d = 2d
The total time taken by the train is t1+t2{t_1} + {t_2}
va=2dt1+t2\therefore {v_a} = \dfrac{{2d}}{{{t_1} + {t_2}}}
=2dd20+d30= \dfrac{{2d}}{{\dfrac{d}{{20}} + \dfrac{d}{{30}}}}
By cancelling the common terms, we will get
va=2110(12+13){v_a} = \dfrac{2}{{\dfrac{1}{{10}}\left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)}}
By rearranging and cross multiplying, we get
=2×102+36= \dfrac{{2 \times 10}}{{\dfrac{{2 + 3}}{6}}}
=20×65= \dfrac{{20 \times 6}}{5}
va=24kmh1\Rightarrow {v_a} = 24km{h^{ - 1}}
Thus, the average speed of the train will be 24kmh124km{h^{ - 1}}

Therefore, from the above explanation, the correct answer is, option (B) 24kmh124km{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