# Question 2b9f0

Aug 23, 2017

color(blue)(rArr(5v_1v_2)/(3v_1+2v_2)

#### Explanation:

Remember that,

$\textcolor{b l u e}{\text{Average speed "=("Total distance travelled")/("Total time taken}}$

The total distance travelled is color(green)(d

Now, let's calculate the time using the formula

$\textcolor{b r o w n}{\text{Speed"=("Distance ")/("Time}}$

When the body moves with velocities ${v}_{1} \mathmr{and} {v}_{2}$, they take time ${t}_{1} \mathmr{and} {t}_{2}$ and cover distances of $\frac{2}{5} d \mathmr{and} \frac{3}{5} d$

So,

rArrcolor(green)(v_1=(2/5 d)/(t_1),t_1=(2/5 d)/v_1

rArrcolor(green)(v_2=(3/5 d)/(t_2),t_2=(3/5 d)/v_2

Now combine all the values

rarr"Average speed"=(d)/((2/5 d)/(v_1)+(3/5 d)/v_2

$\rightarrow \frac{d}{\frac{\left(\frac{2}{5} {\mathrm{dv}}_{2}\right) + \left(\frac{3}{5} {\mathrm{dv}}_{1}\right)}{{v}_{1} {v}_{2}}}$

$\rightarrow \frac{{\mathrm{dv}}_{1} {v}_{2}}{\frac{2}{5} {\mathrm{dv}}_{2} + \frac{3}{5} {\mathrm{dv}}_{1}}$

$\rightarrow \frac{{\cancel{\mathrm{dv}}}_{1} {v}_{2}}{\cancel{d} \left(\frac{2}{5} {v}_{2} + \frac{3}{5} {v}_{1}\right)}$

$\rightarrow \frac{{v}_{1} {v}_{2}}{\frac{2 {v}_{2} + 3 {v}_{1}}{5}}$

color(blue)(rArr(5v_1v_2)/(3v_1+2v_2)#

Hope this helps!!! ☺