Question

A bus travels the first one third distance at a speed of $10km{h^{ - 1}}$, the next one third distance at a speed of $20km{h^{ - 1}}$ and the next one third distance at a speed of $30km{h^{ - 1}}$. Find the average speed of the bus in the whole journey:
A) $20km{h^{ - 1}}$.
B) $\dfrac{{50}}{{11}}km{h^{ - 1}}$.
C) $\dfrac{{180}}{{11}}km{h^{ - 1}}$.
D) $30km{h^{ - 1}}$.

Answer 1

Hint: The average speed is the speed of the body with which if the body have moved instead of three different speeds then it would have taken the exact same time that the body takes with different speeds for different intervals of time. The average speed is defined as the ratio of total distance covered by total time taken.

Formula used:
The formula of the speed is given by,
$ \Rightarrow s = \dfrac{d}{t}$
Where speed is s, the distance is d and the time taken is t.
The formula of the average speed is given by,
$ \Rightarrow {s_{avg.}} = \dfrac{{{d_{total}}}}{{{t_{total}}}}$
Where average speed is ${s_{avg.}}$, the total distance is ${d_{total}}$ and the total time taken is ${d_{total}}$.

Complete step by step solution:
It is given in the problem that a bus travels the first one third distance at a speed of $10km{h^{ - 1}}$, the next one third distance at a speed of $20km{h^{ - 1}}$ and the next one third distance at a speed of $30km{h^{ - 1}}$ and we need to find the average speed of the bus for the whole journey.
For the first journey the bus covers a distance one third of the whole journey and the speed is of $10km{h^{ - 1}}$, let the total distance be d km then the distance for the first journey is $\dfrac{d}{3}$.
The formula of the speed is given by,
$ \Rightarrow s = \dfrac{d}{t}$
Where speed is s, the distance is d and the time taken is t.
The time taken is equal to,
$ \Rightarrow 10 = \dfrac{{\dfrac{d}{3}}}{{{t_1}}}$
$ \Rightarrow {t_1} = \dfrac{d}{{30}}hr.$………eq. (1)
For the second journey we have, one third distance at a speed of $20km{h^{ - 1}}$.
The time taken by the bus will be,
$ \Rightarrow 20 = \dfrac{{\dfrac{d}{3}}}{{{t_2}}}$
$ \Rightarrow {t_2} = \dfrac{d}{{60}}hr.$………eq. (2)
For the third journey we have, one third distance at a speed of $30km{h^{ - 1}}$.
The time taken by the bus will be,
$ \Rightarrow 30 = \dfrac{{\dfrac{d}{3}}}{{{t_2}}}$
$ \Rightarrow {t_2} = \dfrac{d}{{90}}hr.$………eq. (3)
The total time taken will be equal to,
$ \Rightarrow {t_{total}} = {t_1} + {t_2} + {t_3}$
Replacing the equations (1), (2) and (3) in the above relation we get,
$ \Rightarrow {t_{total}} = {t_1} + {t_2} + {t_3}$
$ \Rightarrow {t_{total}} = \dfrac{d}{{30}} + \dfrac{d}{{60}} + \dfrac{d}{{90}}$
$ \Rightarrow {t_{total}} = \dfrac{{6d + 3d + 2d}}{{180}}$
$ \Rightarrow {t_{total}} = \dfrac{{11d}}{{180}}$.
The total distance is $d$.
The formula of the average speed is given by,
$ \Rightarrow {s_{avg.}} = \dfrac{{{d_{total}}}}{{{t_{total}}}}$
Where average speed is ${s_{avg.}}$, the total distance is ${d_{total}}$ and the total time taken is ${d_{total}}$.
$ \Rightarrow {s_{avg.}} = \dfrac{{{d_{total}}}}{{{t_{total}}}}$
$ \Rightarrow {s_{avg.}} = \dfrac{d}{{\dfrac{{11d}}{{180}}}}$
$ \Rightarrow {s_{avg.}} = \dfrac{1}{{\dfrac{{11}}{{180}}}}$
$ \Rightarrow {s_{avg.}} = \dfrac{{180}}{{11}}km{h^{ - 1}}$.

The average speed of the bus is equal to ${s_{avg.}} = \dfrac{{180}}{{11}}km{h^{ - 1}}$. The correct answer is option C.

Note: It is advised to the students to understand and remember the formula of the average speed as it is very helpful in solving problems like these. The formula of the speed can be used to calculate the total time taken.