The co-ordinates of a particle moving in \(x-y\) plane are given by :  \(\mathrm{x}=2+4 \mathrm{t}, \mathrm{y}=3 \mathrm{t}+8 \mathrm{t}^2 .\) The motion of the particle is _______.

\(\frac{d x}{d t}=v_x=4\)

\(\frac{d v_x}{d t}=a_x=0\)

\(y=3 t+8 t^2\)

\(\frac{d y}{d t}=v_y=3+16 t\)

\(\frac{d v_y}{d t}=a_y=16\)

the motion will be uniformly accelerated motion.

For path

\(\mathrm{x}=2+4 \mathrm{t}\)

\(\frac{(\mathrm{x}-2)}{4}=\mathrm{t}\)

Put this value of \(t\) is equation of \(y\)

\(y=3\left(\frac{x-2}{4}\right)+8\left(\frac{x-2}{4}\right)^2\)

this is a quadratic equation so path will be parabola.

Correct answer \((4)\)