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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
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