A particle starts from the origin at t = 0 with an initial velocity of 3.0 ^ i m / s and moves in the x - y plane with a constant acceleration (6.0 ^ i + 4.0 ^ j )m / s ^ 2 . The x - coordinate of the particle at the

Correct answer: C. 60 S _ y = u _ y t + (1)/(2) a _ y t ^ 2 32 = 0 + (1)/(2) × 4 t ^ 2 ⇒ t = 4s S _ x = u _ x t + (1)/(2) a _ x t ^ 2 = 3 × 4 + (1)/(2) × 6 × 16 = 60m

JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2020 · 9 January · Evening Shift · Physics · Motion in Two Dimensions

A particle starts from the origin at t = 0 with an initial velocity...

A particle starts from the origin at

t = 0

with an initial velocity of

3.0\overset{^}{i}m / s and moves in the

x - y

plane with a constant acceleration \left(6.0\overset{^}{i} + 4.0\overset{^}{j}\right)m / {s}^{2} . The x -

coordinate of the particle at the instant when its

y -

coordinate is

32m

D

meters. The value of

D

is:

Options

A $32$
B $50$
C $60$
D $40$

Correct answer

C. $60$

Step-by-step solution

{S}_{y} = {u}_{y}t + \dfrac{1}{2}{a}_{y}{t}^{2}

32 = 0 + \dfrac{1}{2} \times 4{t}^{2} ⇒ t = 4s

{S}_{x} = {u}_{x}t + \dfrac{1}{2}{a}_{x}{t}^{2}

= 3 \times 4 + \dfrac{1}{2} \times 6 \times 16

= 60m

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