Q78 / 90 · JEE Main 2024 (04 Apr Shift 1) · Physics

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

  • Correct answer
    A
    uniformly accelerated having motion along a parabolic path.
  • B
    uniform motion along a straight line.
  • C
    uniformly accelerated having motion along a straight line.
  • D
    non-uniformly accelerated.

Solution

$\begin{aligned} & x=2+4 t \\ & \frac{d x}{d t}=v_x=4 \\ & \frac{d v_x}{d t}=a_x=0 \\ & y=3 t+8 t^2 \\ & \frac{d y}{d t}=v_y=3+16 t \\ & \frac{d v_y}{d t}=a_y=16 \end{aligned}$ 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 $t$ 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.

Hence, the correct answer is option: 'A'.

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