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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 :
non-uniformly accelerated.
uniformly accelerated having motion along a straight line.
uniform motion along a straight line.
uniformly accelerated having motion along a parabolic path.
Solution
$x=2+4 t$
$\frac{d x}{d t}=v_x=4$
$\frac{d v_x}{d t}=a_x=0$
$y=3 t+8 t^2$
$\frac{d y}{d t}=v_y=3+16 t$
$\frac{d v_y}{d t}=a_y=16$
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 $(4)$
