The co-ordinates of a moving particle at any time t are given by `x=ct^(2) and y=bt^(2)`The speed of the particle is

Options

A. `2t(c+b)`
B. `2tsqrt(c^(2)-b^(2))`
C. `tsqrt(c^(2)+b^(2))`
D. `2t sqrt(c^(2)+b^(2))`

Answer

D

Explanation

`|vec(v)|=sqrt(v_(x)^(2)+v_(y)^(2))` here `v_(x)=(dx)/(dt)=2ct, v_(y)=(dy)/(dt)=2bt`
therefore `|vec(v)|=sqrt(4t^(2)(c^(2)+b^(2)))=2tsqrt((c^(2)+b^(2)))`