The coordinate of a particle moving in x-y plane at any time t are (2t,t^2). Find (a) the trajectory of the particle, (b) velocity of particle at time t and (c) acceleration of particle at any time t. please provide solution too! ;0

Hello chanchal

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 ...in vectore form is writtenas 2j   ans

I hope you got the full solution ...any confusion ask furthure

With regards

Yagya

askiitians_expert

(X,Y)=(2t,t^2)

 X=2t and Y=t^2

after solving these two eq we get

 X^2=4Y(parabola)........equation of tragectory 

   any point can be represented as r(vector) = xi+yj 

  r=2ti + t^2j.............1

  V(velocity vector)= dr/dt =2i + 2tj.........2

   magnitude of V =2sqrt(1+t^2)

 a(accleration vector)=dv/dt= 2j............3