The co-ordinate of the particle in x-y plane are given as `x=2+2t+4t^(2)` and `y=4t+8t^(2)`. The motion of the particle is

Options

A. Along a straight line
B. Uniformly accelerated
C. Along a parabolic path
D. Non-uniformly accelerated

Answer

A, B

Explanation

`x=2+2t+4t^(2), y=4t+8t^(2)`
`v_(x)=(dx)/(dt)=2+8t, v_(y)=(dy)/(dt)=4+16t`
`a_(x)=8, a_(y)=16, vec(a)=8hat(i)+16hat(j)=` constant
`y=2(2t+4t^(2)), y=2(x-2)( :' x=2+2t+4t^(2))`
which is the equation of straight line.