JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Kinematics

Starting at time

t = 0

from the origin with speed

1 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

a_{x}

and

a_{y}

, respectively. Then

(A) $a_{x} = 1 ms^{- 2}$ implies that when the particle is at the origin, $a_{y} = 1 ms^{- 2}$ ✓ Correct
(B) $a_{x} = 0$ implies $a_{y} = 1 ms^{- 2}$ at all times✓ Correct
(C)at $t = 0$ , the particle's velocity points in the $x$ -direction✓ Correct
(D) $a_{x} = 0$ implies that at $t = 1$ s, the angle between the particle's velocity and the $x$ axis is $45°$ ✓ Correct

Answer: (A), (B), (C), (D)

Solution

Given:

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

Differentiating:

v_{y} = x v_{x}

Differentiating again:

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

(A) If

a_{x} = 1

at origin (x=0), then

a_{y} = v_{x}^{2} = 1^{2} = 1

✓

(B) If

a_{x} = 0

, then

v_{x}

= constant = 1, so

a_{y} = 1^{2} = 1

at all times ✓

(C) At t=0, x=0, so

v_{y} = 0 \times v_{x} = 0

. Since speed = 1, we have

v_{x} = 1

. Velocity points in x-direction ✓

(D) If

a_{x} = 0

, then

v_{x} = 1

(constant) and

a_{y} = 1

.
At t=1 s:

v_{y} = 0 + a_{y} \times t = 1 \times 1 = 1

tan θ = \frac{v _{y}}{v _{x}} = \frac{1}{1} = 1

, so

θ = 45°

✓

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options