JEE Main Motion in Plane PYQ

1. (JEE Main 2024 (Online) 4th April Morning Shift )

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

The motion of the particle is :

A uniform motion along a straight line. .

B. non-uniformly accelerated.

C. uniformly accelerated having motion along a straight line.

D. uniformly accelerated having motion along a parabolic path.

Correct option is (d)

To determine the nature of the motion of the particle given by its coordinates in the $x$ - $y$ plane, we analyze the given equations for $x$ and $y$ in terms of time $t$ :

$x = 2 + 4 t$
$y = 3 t + 8 t^{2}$

Firstly, the equation for $x$ is of the form $x = x_{0} + v t$ , where $x_{0} = 2$ is the initial position and $v = 4$ is the constant velocity along the $x$ -axis. This suggests a uniform motion along the $x$ -axis because the velocity remains constant with time.

Secondly, the equation for $y$ is a second-degree polynomial in $t$ , which indicates a parabolic path. The presence of the $t^{2}$ term ( $8 t^{2}$ ) signifies acceleration since the position along the $y$ -axis is changing at a rate that itself changes over time.

The equation for $y$ can show two types of motion depending on the terms:

If it was of the form $y = y_{0} + v t$ , it would indicate uniform motion.
If it was of the form $y = y_{0} + v t + \frac{1}{2} a t^{2}$ , where $a$ would represent acceleration, it would indicate uniformly accelerated motion. The presence of the $8 t^{2}$ term here plays a similar role, indicating that the motion is uniformly accelerated in the $y$ -direction due to the constant acceleration implied by this term.

Since the motion in the $y$ direction is determined by a quadratic equation, and the path of the particle depends on both the $x$ and $y$ coordinates, the motion of the particle is not along a straight line but rather follows a parabolic path due to the quadratic (second-degree) dependence on time in the $y$ -coordinate.

Additionally, the acceleration is not changing with time, as deduced from the constant coefficient of the $t^{2}$ term in the $y$ equation, indicating uniform acceleration. Therefore, the motion is uniformly accelerated and follows a parabolic path.

Hence, the correct answer is:

Option D: uniformly accelerated having motion along a parabolic path.

2. (JEE Main 2024 (Online) 27th January Morning Shift )

Position of an ant ( $S$ in metres) moving in $Y$ - $Z$ plane is given by $S = 2 t^{2} \hat{j} + 5 \hat{k}$ (where $t$ is in second). The magnitude and direction of velocity of the ant at $t = 1 s$ will be :

A. $16 m / s$ in $y$ -direction

B. $4 m / s$ in $x$ -direction

C. $9 m / s$ in $z$ -direction

D. $4 m / s$ in $y$ -direction

Correct option is (d)

$\vec{v} = \frac{ds}{dt} = 4 t \hat{j}$

At $t = 1 \sec \vec{v} = 4 \hat{j}$

3. (JEE Main 2022 (Online) 28th July Evening Shift )

At time $t = 0$ a particle starts travelling from a height $7 \hat{z} cm$ in a plane keeping z coordinate constant. At any instant of time it's position along the $\hat{x}$ and $\hat{y}$ directions are defined as $3 t$ and $5 t^{3}$ respectively. At t = 1s acceleration of the particle will be

A. $- 30 \hat{y}$

B. $30 \hat{y}$

C. $3 \hat{x} + 15 \hat{y}$

D. $3 \hat{x} + 15 \hat{y} + 7 \hat{z}$

rrect Option is (B)

$x = 3 t \Rightarrow a_{x} = 0$

$y = 5 t^{3} \Rightarrow a_{y} = 30 t$

$\vec{a} (t = 1) = 30 \hat{y}$

4. (JEE Main 2021 (Online) 20th July Morning Shift )

A butterfly is flying with a velocity $4 \sqrt{2}$ m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :

A. $12 \sqrt{2}$ m

B. 20 m

C. 3 m

D. 15 m

rrect Option is (D)

The given situation can be represented as

JEE Main 2021 (Online) 20th July Morning Shift Physics - Motion Question 85 English Explanation
In the above figure, v₁ is the speed of wind and v₂₁ is the speed of butterfly with respect to wind.

So, v₂₁ can be given as

$v_{21} = 4 \sqrt{2} \cos 45^{\circ} \hat{i} + 4 \sqrt{2} \sin 45^{\circ} \hat{j}$

$= 4 \sqrt{2} \times \frac{1}{\sqrt{2}} \hat{i} + 4 \sqrt{2} \times \frac{1}{\sqrt{2}} \hat{j} = 4 \hat{i} + 4 \hat{j}$

and v₁ can be given as

$v_{1} = - \hat{j}$

$∴$ Velocity of butterfly can be given as

$v_{2} = v_{1} + v_{21} = 4 \hat{i} + 4 \hat{j} - \hat{j} = 4 \hat{i} + 3 \hat{j}$

$∴$ Displacement of butterfly, $D = v_{2} \times t$

$= (4 \hat{i} + 3 \hat{j}) \times 3 = 12 \hat{i} + 9 \hat{j}$

$∴$ Magnitude of displacement, $| D | = \sqrt{12^{2} + 9^{2}} = 15$ m

5. (JEE Main 2021 (Online) 16th March Evening Shift )

A mosquito is moving with a velocity $\vec{v} = 0.5 t^{2} \hat{i} + 3 t \hat{j} + 9 \hat{k}$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?

A. $\tan^{- 1} (\frac{\sqrt{85}}{6})$ from y-axis

B. $\tan^{- 1} (\frac{5}{2})$ from y-axis

C. $\tan^{- 1} (\frac{2}{3})$ from x-axis

D. $\tan^{- 1} (\frac{5}{2})$ from x-axis

rrect Option is (A)

$\vec{v} = (0.5 t^{2} \hat{i} + 3 t \hat{j} + 9 \hat{k})$ m/s

At t = 2 s

$\vec{v} = (2 \hat{i} + 6 \hat{j} + 9 \hat{k})$

Direction cosine along y-axis,

$c o s θ = \frac{(v . \hat{j})}{\sqrt{9^{2} + 6^{2} + 2^{2}}} = \frac{6}{\sqrt{121}} = \frac{6}{11}$

$∴$ $\sin θ = \frac{\sqrt{85}}{11}$

and $\tan θ = \frac{\sqrt{85}}{6}$

$∴$ Mosquito make angle $\tan^{- 1} (\frac{\sqrt{85}}{6})$ from y-axis.