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Question: A bus covers its onward journey with a constant speed \({\ \mathbf{30kmh}}^{-\mathbf{1}}\) and its r...

A bus covers its onward journey with a constant speed  30kmh1{\ \mathbf{30kmh}}^{-\mathbf{1}} and its return journey with a constant speed of 60kmh1{\mathbf{60kmh}}^{-\mathbf{1}} . The average speed for its entire journey is
A) 90kmh1 B) 45kmh1 C) 40kmh1 D) 15kmh1  A) {\ \mathbf{90kmh}}^{-\mathbf{1}}\\\ B) {\ \mathbf{45kmh}}^{-\mathbf{1}} \\\ C) {\ \mathbf{40kmh}}^{-\mathbf{1}} \\\ D) {\ \mathbf{15kmh}}^{-\mathbf{1}}\\\

Explanation

Solution

To solve this question we have to use the relation between average speed, distance travelled and time.Average velocity is defined as the ratio of total displacement of a body to the total time taken by it.

Complete step by step answer:
Let us consider the bus covers a distance (say x) then comes back to the initial point by covering the same distance x. Here it covers the same distance on its onboard journey as well as written journey. As per the question it is given that the velocities are different for onboard journey and return journey, so we can conclude that even if it covers the same distance, the time taken is different for both cases.
Let us say ‘x’ be the distance covered by the bus on its onboard journey and return journey.
v1= 30kmh1{v_1}={\ 30kmh}^{-1} is the velocity of the bus on its onboard journey and v2= 60kmh1{v_2}={\ 60kmh}^{-1} is the velocity of the bus on its return journey. t_1 is the time taken by bus on its onboard journey and t_2 is the time by bus on its return journey.
Here, we have to find average speed throughout the journey. To find that first let us discuss the meaning of the speed. Speed is the rate of change of distance travelled. It can be written mathematically as Speed=distance travelledtime\dfrac{distance\ travelled}{time}
Average speed is the total distance covered by the total time taken by the bus. It can be represented as v=x+xt1+t2v=\dfrac{x+x}{t_1+t_2}.
Further simplifying this equation, we arrive at v=2xt1+t2v=\dfrac{2x}{t_1+t_2} -------(1)
But in this question, all the values are in terms of velocity, so it would be easier for us if we write the equation in terms of velocity itself. To do that we have to eliminate all the other terms.
Since time taken by bus is different in both cases, we try to eliminate the term time and write in terms of velocity and distance.
By using the relation between speed, distance travelled and time we have v=\dfrac{x}{t}
Now we rearrange this equation to get time,t=xv t=\dfrac{x}{v}
Hence,t1=xv1andt2=xv2{t_1}=\dfrac{x}{v_1} and t_2=\dfrac{x}{v_2} Substituting this in equation (1)
we get v=2xxv1+xv2v=\dfrac{2x}{\dfrac{x}{v_1}+\dfrac{x}{v_2}}
Let us simplify this equation further by taking LCM for denominator,
v=2xxv2+xv1v1v2v=\dfrac{2x}{\dfrac{{\rm xv}_2+{\rm xv}_1}{v_1v_2}}
After taking denominator to numerator by reciprocal method and taking out the common term x in the denominator we get
v=2xv1v2x(v2+v1)v=\dfrac{2xv_1v_2}{{x(v}_2+v_1)}
Here x is common in both numerator and denominator. Hence, we can cancel this term, then the equation can be reduced to
v=2v1v2(v2+v1)v=\dfrac{2v_1v_2}{{(v}_2+v_1)} ------ (2)
This equation free from distance and time factors, now we have already the values of velocities we substitute the values in the equation (2)
v=2×30×6030+60 v=360090 v=40kmh1  v=\dfrac{2\times 30\times 60}{30+60} \\\ \Rightarrow v=\dfrac{3600}{90}\\\ \therefore v=40{\rm kmh}^{-1} \\\
Hence, the average speed of the bus is 40kmh140{\rm kmh}^{-1} .The correct answer is option (C).

Note: In most of the cases we may get confused about the word average that if we suppose to find average velocity we just add the velocities and divide it by 2 which is not correct in all the cases. In these kinds of problems, we have to look into the cases if the distance covered will be the same with different velocities then we can use the equation (2) directly instead of doing all the steps.