Find Direction of a Moving Mosquito in 3D – Motion in a Plane JEE PYQ

 

  🧲 Motion in a Plane Physics | Doubtify JEE


💡 Question:

A mosquito is moving with a velocity v = 0.5 t² î + 3t ĵ + 9 k̂ m/s and accelerating in uniform conditions.What will be the direction of motion of the mosquito after 2 seconds?


🖼️ Question Image:

A mosquito is moving with a velocity v = 0.5 t² î + 3t ĵ + 9 k̂ m/s and accelerating in uniform conditions.What will be the direction of motion of the mosquito after 2 seconds?


🔬 Concept Overview:

This is a vector motion problem where you're given a time-dependent velocity vector in three dimensions. To find the direction of motion, we must compute the velocity vector at t = 2 seconds and then determine its direction using unit vector analysis or angle calculation with respect to axes. This is a key concept from the chapter "Motion in a Plane", often asked in JEE Mains or Advanced.


🧠 Step-by-Step Explanation:

Given:

v(t)=0.5t2i^+3tj^+9k^\vec{v}(t) = 0.5t^2 \hat{i} + 3t \hat{j} + 9 \hat{k}

At t=2t = 2 seconds:

  • i-component: 0.5×22=2m/s

  • j-component: 3×2=6m/s3 \times 2 = 6 \, \text{m/s}

  • k-component: 9m/s9 \, \text{m/s}

So,

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

This vector gives us the instantaneous velocity at 2 seconds.


🧭 Finding the Direction:

To find the direction, you can write the unit vector in the direction of velocity:

v=22+62+92=4+36+81=121=11|\vec{v}| = \sqrt{2^2 + 6^2 + 9^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11

So, the unit vector in the direction of motion:

v^=211i^+611j^+911k^\hat{v} = \frac{2}{11} \hat{i} + \frac{6}{11} \hat{j} + \frac{9}{11} \hat{k}

Thus, the mosquito is moving in a 3D direction with the above vector components at 2 seconds. You can also find angles with respect to axes using:

cos(θx)=211,cos(θy)=611,cos(θz)=911\cos(\theta_x) = \frac{2}{11}, \quad \cos(\theta_y) = \frac{6}{11}, \quad \cos(\theta_z) = \frac{9}{11}

This means the direction is closer to the z-axis, moderately inclined to the y-axis, and least toward the x-axis.


✅ Final Answer:

Direction of motion at t = 2 seconds is along the vector:

2i^+6j^+9k^2\hat{i} + 6\hat{j} + 9\hat{k}

or, as a unit vector:

211i^+611j^+911k^\frac{2}{11} \hat{i} + \frac{6}{11} \hat{j} + \frac{9}{11} \hat{k}

🧠 Why this Question is Important:

This is a typical JEE Mains-style vector question that tests your grip on resolving velocity vectors and applying them to real-world motion analysis. These questions enhance your visualization of motion in 3D space, which becomes crucial when solving projectile motion or electromagnetism problems in later chapters.


🧠 Solution (Image):

A mosquito is moving with a velocity v = 0.5 t² î + 3t ĵ + 9 k̂ m/s and accelerating in uniform conditions.What will be the direction of motion of the mosquito after 2 seconds?



🎥 Video Solution:


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