Physics · Motion in one Dimension

JEE Advanced 2020 — Paper 2 — Question 10

Starting at time $\mathrm{t}=0$ from the origin with speed $1 \mathrm{~ms}^{-1}$ , a particle follows a two-dimensional trajectory in the $x-y$ plane so that its coordinates are related by the equation $y=\frac{x^{2}}{2}$ . The $x$ and $y$ components of its acceleration are denoted by $\mathrm{a}_{\mathrm{x}}$ and $\mathrm{a}_{\mathrm{y}}$ , respectively. Then

Option A:
$a_{x}=1 \mathrm{~ms}^{-2}$ implies that when the particle is at the origin, $a_{y}=1 \mathrm{~ms}^{-2}$
Correct
Option B:
$\mathrm{a}_{\mathrm{x}}=0$ implies $\mathrm{a}_{\mathrm{y}}=1 \mathrm{~ms}^{-2}$ at all times
Correct
Option C:
at $t=0$ , the particle's velocity points in the x -direction
Correct
Option D:
$\mathrm{a}_{\mathrm{x}}=0$ implies that at $\mathrm{t}=1 \mathrm{~s}$ , the angle between the particle's velocity and the x axis is $45^{\circ}$
Correct

Answer: A, B, C, D

Step-by-step solution

$y=\frac{x^{2}}{2}$

$v_{y}=x v_{x}$

$a_{y}=x a_{x}+v_{x}^{2}$

$x=0 \Rightarrow a_{y}=a_{x}^{2}=1 \mathrm{~m} / \mathrm{s}^{2}$ , for any value of $a_{x}$

$a_{x}=0 \Rightarrow a_{y}=v_{x}^{2}=1$

$\mathrm{a}_{\mathrm{x}}=0$

$\tan \theta=\frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{v}_{\mathrm{y}}}=\mathrm{x}=1\left(\mathrm{x}=\mathrm{v}_{\mathrm{x}} \mathrm{t}=1\right)$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Physics
Chapter: Motion in one Dimension
Topic: Non-Uniformly Accelerated Motion