Three particles start from the origin at the same time, one with a velocity v1 along the x -axis, second along the negative y -axis with a velocity v2 and third particle moves along the line x=y . The velocity of third particle, so that three may always lie on the same line is :

Question

Three particles start from the origin at the same time, one with a velocity v1 along the x -axis, second along the negative y -axis with a velocity v2 and third particle moves along the line x=y . The velocity of third particle, so that three may always lie on the same line is : (A) ( textv1 + textv2/2) (B) √ textv1 + textv2 (C) ( textv1 textv2/ textv1 + textv2) (D) v=(√2 v1 v2/v2 - v1)

Tardigrade Admin · Accepted Answer

Slope of AC line = Slope of BA line (((vt/√2)-0)/((vt)√2-v1t))=(0 - (- v2 t)/(v1 t - 0)) v=(√2 v1 v2/v2 - v1) Solution

Q. Three particles start from the origin at the same time, one with a velocity $v_{1}$ along the $x$ -axis, second along the negative $y$ -axis with a velocity $v_{2}$ and third particle moves along the line $x = y$ . The velocity of third particle, so that three may always lie on the same line is :

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

$v_{1} + v_{2}$

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

$v = \frac{2 v _{1} v _{2}}{v _{2} - v _{1}}$

Slope of $A C$ line = Slope of $B A$ line
$\frac{( \frac{v t}{2} - 0 )}{( \frac{v t}{2} - v _{1} t )} = \frac{0 - ( - v _{2} t )}{( v _{1} t - 0 )}$
$v = \frac{2 v _{1} v _{2}}{v _{2} - v _{1}}$

Q. Three particles start from the origin at the same time, one with a velocity v1​ along the x -axis, second along the negative y -axis with a velocity v2​ and third particle moves along the line x=y . The velocity of third particle, so that three may always lie on the same line is :

2v1​+v2​​

v1​+v2​​

v1​+v2​v1​v2​​

v=v2​−v1​2​v1​v2​​

Slope of AC line = Slope of BA line (2​vt​−v1​t)(2​vt​−0)​=(v1​t−0)0−(−v2​t)​ v=v2​−v1​2​v1​v2​​

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

$v_{1} + v_{2}$

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

$v = \frac{2 v _{1} v _{2}}{v _{2} - v _{1}}$

Slope of $A C$ line = Slope of $B A$ line
$\frac{( \frac{v t}{2} - 0 )}{( \frac{v t}{2} - v _{1} t )} = \frac{0 - ( - v _{2} t )}{( v _{1} t - 0 )}$
$v = \frac{2 v _{1} v _{2}}{v _{2} - v _{1}}$