Q122 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Physics

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.

  • A
    $\frac{72}{5} \mathrm{~ms}^{-1}$
  • B
    $\frac{36}{3} \mathrm{~ms}^{-}$
  • Correct answer
    C
    $\frac{72}{3} \mathrm{~ms}^{-1}$
  • D
    $\frac{36}{5} \mathrm{~ms}^{-1}$

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 $$ \begin{aligned} v_{a v} & =\frac{2 \times 18 \times 36}{18+36} \\ & =\frac{2 \times 36 \times 18}{54}=\frac{72}{3} \mathrm{~m} / \mathrm{s} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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