The coordinates of a particle moving in a plane are given by x = acos pt and y = bsin pt where a, b (< a) and p are positive constants of appropriate dimensions. Then,

x=acos⁡pt⇒cos⁡pt=xa    ........(1)y=bsin⁡pt⇒sin⁡pt=yb      .........(2)Squaring and adding Eqs . (1) and (2), we get                   x2a2+y2b2=1Therefore , path of the particle is an ellipse . Now  r=acos⁡pti^+bsin⁡ptj^          v=drdt=−apsin⁡pti^+bpcos⁡ptj^          a=dvdt=−ap2cos⁡pti^−bp2sin⁡ptj^we can see that at t=π/2p,v⋅a=0or velocity is perpendicular to aSimilarly,   a=−p2(r)i.e. acceleration of particle is always directed towards origin .Further at time t = 0 particle is at (a, 0) and at time t=π2p particle is at (0, b).So, distance travelled is greater than a.