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# A car moves north at a speed of $54km{h^{ - 1}}$ for $1hr$. Then it moves eastward with the same speed for the same duration. The average speed and average velocity of car for complete journey is(A) $54km{h^{ - 1}},0$(B) $15m{s^{ - 1}},\dfrac{{15}}{{\sqrt 2 }}m{s^{ - 1}}$(C) $0,0$(D) $0,\dfrac{{54}}{{\sqrt 2 }}m{s^{ - 1}}$

Last updated date: 15th Jun 2024
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Hint: Here we are using the concept of average speed and average velocity to solve this problem. For that you need to know what is average speed and average velocity.
Average speed: It can be explained as the total distance travelled by a car or bike or any vehicle for a particular time interval and we divide the total distance travelled by total time taken.
Average velocity: It is calculated as the total displacement of the object in that time interval divided by total time taken.

Complete step by step solution:

First of all we need to know which formula we are going to use here to solve this problem.
${\text{Average speed}}\left( {{v_{av}}} \right) = \dfrac{{{\text{Total distance travelled}}}}{{{\text{Total time taken}}}}$
$\Rightarrow {V_{av}} = \dfrac{{54 + 54}}{2}$
$\Rightarrow v = 54km{h^{ - 1}}$
$\Rightarrow {v_{av}} = 15m{s^{ - 1}}$
${\text{Average Velocity}}\left( {{{\vec v}_{av}}} \right) = \dfrac{{{\text{Total displacement}}}}{{{\text{Total time taken}}}}$
$\Rightarrow {v_{av}} = \dfrac{{\sqrt {{{\left( {54} \right)}^2} + {{\left( {54} \right)}^2}} }}{2}$
$\Rightarrow {\vec v_{av}} = \dfrac{{54\sqrt 2 }}{2}$
$\Rightarrow {v_{av}} = \dfrac{{54}}{{\sqrt 2 }}km{h^{ - 1}}$
$\Rightarrow {v_{av}} = \dfrac{{15}}{{\sqrt 2 }}m{s^{ - 1}}$

The correct answer is option B