Question

A particle is moving in a plane with velocity given by $\overrightarrow{v} = \widehat{i}u_{0} + \widehat{j}$ aωcosωt if the particle is at origin at t = 0. Distance from origin at time 3π/2ω is

• A. $\sqrt{a^{2} + \left( 3\pi u_{0}/2\omega \right)^{2}}$
• B. $\sqrt{a^{2} + \left( 2\pi u_{0}/\omega \right)^{2}}$
• C. $\left( \pi u_{0}/\omega \right)^{2} + a^{2}$
• D. $\sqrt{a^{2} + \left( 2\pi u_{0}/3\omega \right)^{2}}$

Answer 1

Answer: (A)

$\sqrt{a^{2} + \left( 3\pi u_{0}/2\omega \right)^{2}}$

Answer 2

Comparing the given equation with $\overrightarrow{v}$= $\widehat{i}$v_x + $\widehat{j}$v_y, we get v_x = u₀ and dy/dt = aω cos ωt or dx/dt =u₀ and dy/dt = aωcosωt. Integrating x = $\int_{}^{}{u_{0}dt}$ and y = $\int_{}^{}{a\omega}$ cos ωdt or x = u₀t + c₁ and y = 0 we get c₁ and c₂ as zero ∴ x = u₀ t and y = a sin ωt but t = 3π/2ω ∴ x = u₀(3π/2ω) and y = -a. Then distance from origin, d = $\sqrt{x^{2} + y^{2}} = \sqrt{a^{2} + \left( 3\pi u_{0}/2\omega \right)^{2}}$