Question

A particle starts from the origin at t = 0. It moves in a plane with a velocity given by: $\overrightarrow{v} = v_{0}\widehat{i}$+ (awcoswt)$\widehat{j}$. The equation of trajectory of the particle is –

• A. y = a sin wt
• B. y = a coswt
• C. y = a sin$\left( \frac{\omega x}{v_{0}} \right)$
• D. y = a cos$\left( \frac{\omega x}{v_{0}} \right)$

Answer 1

Answer: (C)

y = a sin$\left( \frac{\omega x}{v_{0}} \right)$

Answer 2

v_x = $\frac { \mathrm { dx } } { \mathrm { dt } }$ = v₀ ̃ x = v₀t, v_y = $\frac { \mathrm { dy } } { \mathrm { dt } }$ = aw cos wt

̃ y = $\int \mathrm { a } \omega$ cos wt dt = $\frac { \mathrm { a } \omega \sin \omega \mathrm { t } } { \omega }$ = a sin wt = a sin w $\left( \frac { \mathrm { x } } { \mathrm { v } _ { 0 } } \right)$