Question

A particle is moving in xy-plane with a constant speed $v_0$ such that its y displacement is given by $y = \alpha e^{-\frac{2v_x}{\sqrt{3}v_0}}$, where $v_x$ 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 $-\frac{1}{\sqrt{3}}$, then the displacement of the particle in y-direction at the instant is

• A. $\alpha e^{-1}$
• B. $\alpha e^{-2}$
• C. Zero
• D. $\alpha^2 e$

Answer 1

Answer: (A)

$\alpha e^{-1}$

Answer 2

1. The slope of the tangent is

   $$\frac{dy}{dx} = \frac{v_y}{v_x} = -\frac{1}{\sqrt{3}},$$

   so $v_y = -\frac{v_x}{\sqrt{3}}$.

2. With constant speed $v_0$, we have

   $$v_x^2 + v_y^2 = v_0^2.$$

   Substituting $v_y = -\frac{v_x}{\sqrt{3}}$ gives:

   $$v_x^2 + \frac{v_x^2}{3} = v_0^2 \quad \Longrightarrow \quad \frac{4v_x^2}{3} = v_0^2.$$

   Thus,

   $$v_x = \frac{\sqrt{3}}{2}\,v_0 \quad (\text{since } v_x>0).$$

3. The y-displacement is given by

   $$y=\alpha \exp\Bigl[-\frac{2v_x}{\sqrt{3}v_0}\Bigr].$$

   Substituting $v_x = \frac{\sqrt{3}}{2}\,v_0$:

   $$y=\alpha\exp\Bigl[-\frac{2\left(\frac{\sqrt{3}}{2}v_0\right)}{\sqrt{3}v_0}\Bigr] = \alpha\exp(-1).$$