(A) Given that, speed of train from city-$A$ to city-$B$ is $v_1 = 18 \,ms^{-1}$.Speed of train from city-$B$ to city-$A$ is $v_2 = 36 \,ms^{-1}$.Sinc

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(A) Given that, speed of train from city-$A$ to city-$B$ is $v_1 = 18 \,ms^{-1}$.Speed of train from city-$B$ to city-$A$ is $v_2 = 36 \,ms^{-1}$.Since the distance travelled from $A$ to $B$ and $B$ to $A$ is the same, we use the formula for average speed over equal distances:$v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$Substituting the values:$v_{av} = \frac{2 \times 18 \times 36}{18 + 36}$$v_{av} = \frac{2 \times 18 \times 36}{54}$$v_{av} = \frac{1296}{54} = 24 \,ms^{-1}$.

Class 11 Physics Motion in Straight Line Average Velocity and Average Speed

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$A$ train travels from city-$A$ to city-$B$ with a constant speed of $18 \,ms^{-1}$ and returns back to city-$A$ with a constant speed of $36 \,ms^{-1}$. Find its average speed during the journey. (in $\,ms^{-1}$)

AP EAMCET 2021 All AP EAMCET

A
$24$
B
$12$
C
$27$
D
$24$

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Motion in Straight Line

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Average Velocity and Average Speed

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$A$ train travels from city-$A$ to city-$B$ with a constant speed of $18 \,ms^{-1}$ and returns back to city-$A$ with a constant speed of $36 \,ms^{-1}$. Find its average speed during the journey. (in $\,ms^{-1}$)

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