Question

A moving body travels distance d at uniform speed v1 and then travels distance d at uniform speed v2. Prove that its average speed is (2*v1*v2)÷(v1 +v2)

Answer 1

A moving body travels distance d at uniform speed $v_1$ and then travels distance d at uniform speed $v_2$.

time taken by body to travel d distance with speed $v_1$, t = d/$v_1$

also time taken by body to travel d distance with speed $v_2$, t' = d/$v_2$

total distance travelled by body , s = d + d = 2d

and total time taken , T = t + t'

so, average speed = 2d/(t + t')

= 2d/(d/$v_1$ + d/$v_2$)

= 2d/d{1/$v_1$ + 1/$v_2$}

= 2/{($v_1$ + $v_2$)/$v_1$.$v_2$}

= $\frac{2v_1v_2}{v_1+v_2}$ hence, proved