If a car covers $2 / 5^{t h}$ of the total distance with $v_{1}$ speed and $3 / 5^{t h}$ distance with $v_{2}$ then average speed is

$\frac{1}{2} \sqrt{v_{1} v_{2}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{v_{1} + v_{2}}{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{v_{1} v_{2}}{3 v_{1} + 2 v_{2}}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$\frac{v_{1} v_{2}}{3 v_{1} + 2 v_{2}}$

$A v e r a g e s p e e d = \frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n}$
= $\frac{x}{\frac{2 x / 5}{v_{1}} + \frac{3 x / 5}{v_{2}}} = \frac{5 v_{1} v_{2}}{3 v_{1} + 2 v_{2}}$

Suggest Corrections

Similar questions

\frac{2}{5^{t h}}

v_{1}

\frac{3}{5^{t h}}

v_{2}

\frac{2}{5^{t h}}

v_{1}

\frac{3}{5^{t h}}

v_{2}

Q : A car moves2/5^th of distance with v₁ velocity and left of 3/5^th distance with v₂ velocity find its average speed?

If a car covers 2/5th of the total distance with v₁speed and 3/5th of the total distance with v₂ speed, then average speed of the car is

a moving body travels the distance 's' with constant speed v1 distance 's' with uniform speed v2 prove that its average speed is 2v1 v2 by v1+v2

Join BYJU'S Learning Program