Question 24 of 308

A car travels a distance of $x$ with speed ${v}_{1}$ and then same distance $x$ with speed ${v}_{2}$ in the same direction. The average speed of the car is:

A$\dfrac{{v}_{1}{v}_{2}}{2\left({v}_{1}+{v}_{2}\right)}$B$\dfrac{{v}_{1}+{v}_{2}}{2}$C$\dfrac{2x}{{v}_{1}+{v}_{2}}$D$\dfrac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}$

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Answer: (D) $\dfrac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}$

Average velocity of car is:
${v}_{avg}=\dfrac{\text{Total displacement}}{\text{Total time}}$
As the car is travelling along a straight line without turning, the magnitude of total displacement is same as the distance covered.
The time taken to cover the first half is $\dfrac{x}{{v}_{1}}$ and the time taken to cover the second half is $\dfrac{x}{{v}_{2}}$
$∴{v}_{avg}=\dfrac{2x}{\dfrac{x}{{v}_{1}}+\dfrac{x}{{v}_{2}}}$
$\Rightarrow {v}_{avg}=\dfrac{2x\left({v}_{1}{v}_{2}\right)}{{v}_{1}x+{v}_{2}x}=\dfrac{2{v}_{1}{v}_{2}}{{v}_{1}+{v}_{2}}$

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