A body covers ((2/5))th of the total distance with speed v1 and ((3/5))th with v2 . The average speed of the body will be

Question

A body covers ((2/5))th of the total distance with speed v1 and ((3/5))th with v2 . The average speed of the body will be (A) (5 v1 v2/3 v1 + 2 v2) (B) (v1 + v2/2) (C) (1/2)√v1 v2 (D) (2 v1 v2/v1 + v2)

Tardigrade Admin · Accepted Answer

Let the total distance is x, then we have t1=(2/5) (x/v1) and t2=(3/5) (x/v2) Since, average speed, va v=( texttotal distance/ texttotal time)=(x/t1 + t2) =(x/(2 x)5 v1 + (3 x)5 v2=(5 v1 v2/3 v1 + 2 v2) ∴ v/a v)=(5 v1 v2/3 v1 + 2 v2)

Q. A body covers $(\frac{2}{5})^{t h}$ of the total distance with speed $v_{1}$ and $(\frac{3}{5})^{t h}$ with $v_{2}$ . The average speed of the body will be

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

$\frac{v _{1} + v _{2}}{2}$

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

$\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$

Q. A body covers (52​)th of the total distance with speed v1​ and (53​)th with v2​ . The average speed of the body will be

3v1​+2v2​5v1​v2​​

2v1​+v2​​

21​v1​v2​​

v1​+v2​2v1​v2​​

Let the total distance is x, then we have t1​=52​v1​x​ and t2​=53​v2​x​ Since, average speed, vav​=total timetotal distance​=t1​+t2​x​ =5v1​2x​+5v2​3x​x​=3v1​+2v2​5v1​v2​​ ∴vav​=3v1​+2v2​5v1​v2​​

Q. A body covers $(\frac{2}{5})^{t h}$ of the total distance with speed $v_{1}$ and $(\frac{3}{5})^{t h}$ with $v_{2}$ . The average speed of the body will be

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

$\frac{v _{1} + v _{2}}{2}$

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

$\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$