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Question

(D) Let the total distance be $x$.Time taken to cover the first part of the distance $(d_1 = \frac{2}{5}x)$ with speed $v_1$ is $t_1 = \frac{d_1}{v_1}

Vedclass · Accepted Answer

$\frac{5v_1 v_2}{3v_1 + 2v_2}$

Vedclass · Answer

(D) Let the total distance be $x$.Time taken to cover the first part of the distance $(d_1 = \frac{2}{5}x)$ with speed $v_1$ is $t_1 = \frac{d_1}{v_1} = \frac{2x}{5v_1}$.Time taken to cover the second part of the distance $(d_2 = \frac{3}{5}x)$ with speed $v_2$ is $t_2 = \frac{d_2}{v_2} = \frac{3x}{5v_2}$.Average speed is defined as the total distance divided by the total time taken:$	ext{Average speed} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{x}{t_1 + t_2}$Substituting the values of $t_1$ and $t_2$:$	ext{Average speed} = \frac{x}{\frac{2x}{5v_1} + \frac{3x}{5v_2}}$$	ext{Average speed} = \frac{x}{\frac{2xv_2 + 3xv_1}{5v_1 v_2}}$$	ext{Average speed} = \frac{5v_1 v_2}{3v_1 + 2v_2}$

$A$ car covers $2/5^{th}$ of the total distance with speed $v_1$ and $3/5^{th}$ of the total distance with speed $v_2$. What is its average speed?

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