Question

An aeroplane covers 2500km, 1200km and 500km at the rate of 500km/hr, 400km/hr and 250km/hr respectively. Find its average speed for the whole journey.

Answer 1

Hint: Average speed of a body is defined as the total distance covered by the body divided by the total time taken by the body. Calculate the total distance covered by the aeroplane. Then calculate the time taken in covering each distance and sum them to find the total time taken. Divide the two to calculate the average speed of the aeroplane.

Formula used:
$v=\dfrac{d}{t}$

Complete answer:
Average speed of a body is defined as the total distance covered by the body divided by the total time taken by the body.
Suppose that a body covers a total distance d in a time interval t, then its average speed is $v=\dfrac{d}{t}$.
It is given that aeroplane covers a route with three different velocities. First, it covers a distance of 2500km with a speed of 500km/hr, then a distance of 1200km with a speed of 400km/hr. Finally, it covers a distance of 500km with a speed of 250km/hr.
Let the three distances be ${{d}_{1}},{{d}_{2}},{{d}_{3}}$ respectively. And the corresponding three speeds be ${{v}_{1}},{{v}_{2}},{{v}_{3}}$
Then the total distance covered by the aeroplane is ${{d}_{1}}+{{d}_{2}}+{{d}_{3}}$.
Let us calculate the total time taken. For that, let us first take the time taken to cover each of the distances.
Let the time taken to covers ${{d}_{1}},{{d}_{2}},{{d}_{3}}$ be ${{t}_{1}},{{t}_{2}},{{t}_{3}}$ respectively.
Now we will use the formula $t=\dfrac{d}{v}$.
This means that ${{t}_{1}}=\dfrac{{{d}_{1}}}{{{v}_{1}}}$, ${{t}_{2}}=\dfrac{{{d}_{2}}}{{{v}_{2}}}$ and ${{t}_{3}}=\dfrac{{{d}_{3}}}{{{v}_{3}}}$.
Hence, the total time taken by the aeroplane is ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}=\dfrac{{{d}_{1}}}{{{v}_{1}}}+\dfrac{{{d}_{2}}}{{{v}_{2}}}+\dfrac{{{d}_{3}}}{{{v}_{3}}}$.
Therefore, the average speed of the aeroplane is $v=\dfrac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}}=\dfrac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}}{\dfrac{{{d}_{1}}}{{{v}_{1}}}+\dfrac{{{d}_{2}}}{{{v}_{2}}}+\dfrac{{{d}_{3}}}{{{v}_{3}}}}$.
Substitute the values of the distances and the speeds.
$\Rightarrow v=\dfrac{2500+1200+500}{\dfrac{2500}{500}+\dfrac{1200}{400}+\dfrac{500}{250}}=\dfrac{4200}{5+3+2}=\dfrac{4200}{10}=420km{{h}^{-1}}$.
Therefore, the average speed of the aeroplane is 420km/hr.

Note:
One of the common mistakes that a student may do is that the students may think that the average speed of the body will be the average of the three speeds. However, it is a false assumption.
We can also prove it by taking the mean of the three speeds.
The mean of the speeds will be $\dfrac{{{v}_{1}}+{{v}_{3}}+{{v}_{3}}}{3}=\dfrac{500+400+250}{3}=\dfrac{1150}{3}=383.33km/hr$.
This value is not equal to the average speed that we calculated.