Question

If a car covers $2/5^{th}$ of the total distance with $v_1$ speed and $3/5^{th}$ distance with $v_2$ speed, then average speed of the car is

• A. $\frac{1}{2}\sqrt{v_1{v}_2}$
• B. $\frac{v_1+v_2}{2}$
• C. $\frac{2v_1{v}_2}{v_2+v_1}$
• D. $\frac{5v_1{v}_2}{2v_2+3v_1}$

Answer 1

The correct option is D $\frac{5v_1{v}_2}{2v_2+3v_1}$

Let total distance covered by the car be $d$
So, we have, average speed as
$v_{avg}=\frac{\text{Total distance}}{\text{Total time}}$
So, for $2/5^{th}$ distance, time taken is given by
$t_1=\frac{2d}{5v_1}$
Similarly, for $3/5^{th}$ distance, time taken is given by
$t_2=\frac{3d}{5v_2}$
So, total time taken is $t_1+t_2=\frac{2d}{5v_1}+\frac{3d}{5v_2}=\left(\frac{2v_2+3v_1}{5v_1{v}_2}\right)d$
Thus, average speed is given by
$v_{avg}=\frac{d}{\left(\frac{2v_2+3v_1}{5v_1{v}_2}\right)d}=\frac{5v_1{v}_2}{2v_2+3v_1}$