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Chanchal Kumar Grade: 11
        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
6 years ago

Answers : (2)

Askiitians_Expert Yagyadutt
askIITians Faculty
74 Points
										

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

6 years ago
vikas askiitian expert
510 Points
										

(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           


 


 


 

6 years ago
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