Kinematics

JEE Main level, single correct, medium

215 of 308

Question 215 of 308

The co-ordinates of a particle moving in x−yx-y plane are given by : x=2+4t,y=3t+8t2x=2+4 \mathrm{t}, y=3 \mathrm{t}+8 \mathrm{t}^2.
The motion of the particle is :

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Answer: (A) uniformly accelerated having motion along a parabolic path.

$\begin{aligned}
& x=2+4 t \\
& \frac{d x}{d t}=v_x=4 \\
& \frac{d v_x}{d t}=a_x=0 \\
& y=3 t+8 t^2 \\
& \frac{d y}{d t}=v_y=3+16 t \\
& \frac{d v_y}{d t}=a_y=16
\end{aligned}$
the motion will be uniformly accelerated motion.
For path
$\begin{aligned}
& \mathrm{x}=2+4 \mathrm{t} \\
& \frac{(\mathrm{x}-2)}{4}=\mathrm{t}
\end{aligned}$
Put this value of tt is equation of yy
y=3(x−24)+8(x−24)2y=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.

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