The correct options are
A the path of the particle is an ellipse
B the velocity and acceleration of the particle are normal to each other at
t=π/(2p) C the acceleration of the particle is always directed towards a focus
x=a cos pt ⇒ cos(pt)=xa ...(i)
y=b sin pt ⇒ sin(pt)=yb ...(ii)
Squaring and adding (i) and (ii), we get
x2a2+y2b2=1 ∴ Path of the particle is in ellipse.
Hence option (a) is correct.
From the given equations
We can find,
dxdt=vx=−ap sin pt d2x=ax=−ap2 cos pt dydt=vy=bp cos pt and
d2ydt=ay=−bp2 sin pt At time
t=π2p or pt = π2 ax and vy become zero (because cos
π2=0) only vx and ay are left. or we can say that velocity is along negative x-axis and acceleration along y-axis.
Hence, at
t=π2p, velocity and acceleration force the particle are normal to each other. So option (b) is also correct. At t = t, position of the particle
→r(t)=x^i+y^j= a cos pt^i+b sin pt^j and acceleration of the particle is
→a(t)=ax^i+ay^j=p2[a cos pt^i+b sin pt^j] =p2[x^i+y^j=−p2→r(t)] Therefore acceleration of the particle is always directed towards origin.
Hence option (c) is also correct.
At t = 0, particle is at (a, 0) and at
t=π2p, particle is at (0, b). Therefore, the distance covered is one fourth of the elliptical path not a.
Hence option (d) is wrong.