A particle is moving in xy-plane with a constant speed v0 such that its y displacement is given by y=α e- (2 vx/√3 v0), where vx is component of velocity along the x- axis. If at some instant x component of it velocity is positive and the slope of the tangent on its path is -(1/√3), then the displacement of the particle in y- direction at the instant is

Question

A particle is moving in xy-plane with a constant speed v0 such that its y displacement is given by y=α e- (2 vx/√3 v0), where vx is component of velocity along the x- axis. If at some instant x component of it velocity is positive and the slope of the tangent on its path is -(1/√3), then the displacement of the particle in y- direction at the instant is (A) α e- 1 (B) α e- 2 (C) Zero (D) α 2e

Tardigrade Admin · Accepted Answer

Here, particle is moving in two - dimensional

(x - y)

plane, with constant speed so slope of
graph

\frac{d y}{d x} = \frac{v _{y}}{v _{x}}

and

v_{0} = v_{x}^{2} + v_{y}^{2}

.
As slope,

\frac{d y}{d x} = - \frac{1}{3} = \frac{v _{y}}{v _{x}}

So,

v_{x} = - v_{y} 3

On squaring and adding

v_{x}^{2} + v_{y}^{2} = v_{0}^{2}

3 v_{y}^{2} + v_{y}^{2} = v_{0}^{2}

v_{y} = \pm \frac{v _{0}}{2}

v_{x} = \mp \frac{3 v _{0}}{2}

So,

y = α e^{\frac{- 2}{3 v _{0}}} \times \frac{3 v _{0}}{2}

y = α e^{- 1}

$α e^{- 1}$

$α e^{- 2}$

Zero

$α^{2} e$

αe−1

αe−2

Zero

α2e

$α e^{- 1}$

$α e^{- 2}$

$α^{2} e$