1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The velocity of a particle moving in the x−y plane is given by dxdt=4πsin(3πt) and dydt=3πcos(3πt). If the particle starts from origin at t=0 then the path of the particle is a/an

A
straight line
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
hyperbola
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
ellipse
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
parabola
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C ellipse For motion along x− axis, dxdt=4πsin(3πt) ⇒dx=4πsin(3πt)dt On integrating it with limit, x=0 to x and t=0 to t, we get, x=43[1−cos(3πt)] ⇒cos(3πt)=1−3x4 .............(1) For motion along y− axis, dydt=3πcos(3πt) ⇒dy=3πcos(3πt)dt On integrating it with limit, y=0 to y and t=0 to t, we get, y=sin(3πt) ⇒sin(3πt)=y .............(2) We know that, sin2θ+cos2θ=1. From (1) and (2), y2+(1−3x4)2=1 This equation is similar to the equation of an ellipse, x2A2+y2B2=1

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program