A train travels from city- $A$ to city- $B$ with a constant speed of $18 \mathrm{~ms}^{-1}$ and returns back…
A train travels from city- $A$ to city- $B$ with a constant speed of $18 \mathrm{~ms}^{-1}$ and returns back to city- $A$ with a constant speed of $36 \mathrm{~ms}^{-1}$. Find its average speed during the journey.
$\frac{72}{5} \mathrm{~ms}^{-1}$
$\frac{36}{3} \mathrm{~ms}^{-}$
$\frac{72}{3} \mathrm{~ms}^{-1}$
$\frac{36}{5} \mathrm{~ms}^{-1}$
Solution
Given that, speed of train from city- $A$ to city- $B$,
$
v_1=18 \mathrm{~m} / \mathrm{s}
$
Speed of train from city- $B$ to city- $A$
$
v_2=36 \mathrm{~m} / \mathrm{s}
$
In this process, the distance travelled by train from $A$ to $B$ and $B$ to $A$ will be equal.
Now, using the expression of average speed for equal distance, we know
$
v_{a v}=\frac{2 v_1 v_2}{v_1+v_2}
$
By substituting the values, we get
$
\begin{aligned}
v_{a v} & =\frac{2 \times 18 \times 36}{18+36} \\
& =\frac{2 \times 36 \times 18}{54}=\frac{72}{3} \mathrm{~m} / \mathrm{s}
\end{aligned}
$