Question

The coordinates of a particle moving in xy plane at any instant of time t are x=4t square,y=3t square. the speed of particle at that instant is

Answer 1

a) Given co-ordinate of the particle as ( 2t,t^2)

means at ny time t....x = 2t and y = t^2

i.e t = x/2 and so y = ( x/2)^2 or 4y = x^2

Hence the trajectory of the particle is a parabola with equation x^2 = 4y

b) x = 2t dx/dt = 2

y=t^2 dy/dt = 2t

So velocity at any time is (2,2t) or in vector form 2i+ j.(2t)

c) x = 2t .dx^2/dt^2 = 0

y = t^2 .dy^2/dt^2 = 2

so acceleration at any time is (0,2) and 2j in vectors

Answer 2

r = 2t

θ = 4t

Path length = rθ = (2t)(4t) = $8t^{2}$

t = 1s

Speed = Path length/t = $8t^{2}$/1

= 8*1*1

=8m/s