Correct option is : (4) uniformly accelerated having motion along a parabolic path.
\(\mathrm{x}=2+4 \mathrm{t}\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{x}}=4\)
\(\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\mathrm{a}_{\mathrm{x}}=0\)
\(\mathrm{y}=3 \mathrm{t}+8 \mathrm{t}^{2}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{y}}=3+16 \mathrm{t}\)
\(\frac{d v_{y}}{d t}=a_{y}=16\)
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.
