A particle of mass m movies along a curve `y=x^(2)`. When particle has x-co-ordinate as `1//2` and x-component of velocity as `4 m//s`. Then :-

Options

A. The position coordinate of particle are `(1//2, 1//4)`
B. The velocity of particle will be along the line `4x-4y-1=0`
C. The magnitude of velocity at that instant is `4sqrt(2) m//s`
D. The magnitude of angular momentum of particle about origin at that position is 0.

Answer

A, B, C

Explanation

`y=x^(2), y_(x=1/2)=1/4, (dy)/(dt)=2x(dx)/(dt)=2x v_(x)`
`v_(y)=2xx1/2xx4(at x=1/2, v_(x)=4)`
`v_(y)=4m//s, vec(v)_(x=1/2)=4hat(i)+4hat(j), |vec(v)|=4sqrt(2)`
Slope of line `4x-4y-1=0` is `tan 45^(@)=1` and also the slope of velocity is 1.