A particle moves along the positive branch of the curve `y = (x^(2))/(2)` where `x = (t^(2))/(2),x` and y are measured in metres and t in second. At `t = 2s`, the velocity of the particle is

Options

A. `(2hati-4hatj)m//sec`
B. `(2hati+4hatj)` m/sec
C. `(2hati+2hatj)` m/sec
D. `(4hati-2hatj)` m/sec

Answer

b

Explanation

`x=(t^(2))/(2)rArrv_(x)=(dx)/(dt)=t`
`y=(x^(2))/(2)=(t^(4))/(8)rArrv_(y)=(dy)/(dt)=(1)/(2)t^(3)`
when t=2s, `v_(x)=2` and `v_(y)=4`
`therefore vecv=2hati+4hatj`