Physics, asked by lewarends, 1 year ago

If x and y coordinates of a particle moving in xy plane at some instant of time are x = t^{2} and y = 3t^{2} then:


1) The particle has tangential acceleration only


2) The particle has centripetal acceleration only


3) Net acceleration of the particle is zero


4) The particle has both centripetal and tangential acceleration




Please explain your answer :)

Answers

Answered by Anonymous
1373

Answer:

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

Answered by SwaggerGabru
1

Answer:

If x and y coordinates of a particle moving in xy plane at some instant of time are x = t^{2} and y = 3t^{2} then:

1) The particle has tangential acceleration only

2) The particle has centripetal acceleration only

Explanation:

If x and y coordinates of a particle moving in xy plane at some instant of time are x = t^{2} and y = 3t^{2} then:

1) The particle has tangential acceleration only

2) The particle has centripetal acceleration only

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