A particle of mass m moves along a curve `y=x^2`. When particle has x-coordinate as `(1)/(2)` m and x-component of velocity as `4(m)/(s)`, then

Options

A. the position coordinate of particle are `((1)/(2)`,`(1)/(4))`m
B. The velocity of particle will be along the line `4x-4y-1=0`.
C. the magnitude of velocity at that instant is `4sqrt2(m)/(s)`
D. The magnitude of velocity at that instant is `2sqrt2(m)/(s)`.

Answer

A, B, C

Explanation

On the curve `y=x^2` at `x=(1)/(2)impliesy=(1)/(4)`
Hence the coordinate `((1)/(2)`,`(1)/(4))`
Differentiation: `y=x^2`, `v_y=2xv_x`
`v_y=2((1)/(2))(4)=4(m)/(s)`
Which satisfies the line
`4x-4y-1=0` (tangent to the curve) and magnitude of velocity:
`|vecv|=sqrt(v_x^2+v_y^2)=4sqrt2` m/s