If x and y coordinates of a particle moving in xy plane at some instant of time are x = and y = 3 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
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
Answer:
If x and y coordinates of a particle moving in xy plane at some instant of time are x = and y = 3 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 = and y = 3 then:
1) The particle has tangential acceleration only
2) The particle has centripetal acceleration only