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Motion in A Plane question

2021 · 16 Mar · Shift 2 · Q51
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Motion in A Plane question

2021 · 16 Mar · Shift 2 · Q51

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A mosquito is moving with a velocity v→=0.5t2i^+3tj^+9k^\overrightarrow v = 0.5{t^2}\widehat i + 3t\widehat j + 9\widehat kv=0.5t2i+3tj​+9k m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?
  1. A
    tan⁡−1(856){\tan ^{ - 1}}\left( {{\sqrt {85} } \over 6}\right)tan−1(685​​) from y-axis
  2. B
    tan⁡−1(52){\tan ^{ - 1}}\left( {{5 \over 2}} \right)tan−1(25​) from y-axis
  3. C
    tan⁡−1(23){\tan ^{ - 1}}\left( {{2 \over 3}} \right)tan−1(32​) from x-axis
  4. D
    tan⁡−1(52){\tan ^{ - 1}}\left( {{5 \over 2}} \right)tan−1(25​) from x-axis
View written solutionFree

Correct answer: A

  1. The velocity of the mosquito is given by
v⃗=0.5t2 i^+3t j^+9 k^\vec v = 0.5t^2\,\hat i + 3t\,\hat j + 9\,\hat kv=0.5t2i^+3tj^​+9k^

We need the direction of motion after 2 s2\text{ s}2 s, so evaluate the velocity at t=2t=2t=2.

  1. Substitute t=2t=2t=2:
vx=0.5(2)2=2,vy=3(2)=6,vz=9v_x = 0.5(2)^2 = 2, \qquad v_y = 3(2) = 6, \qquad v_z = 9vx​=0.5(2)2=2,vy​=3(2)=6,vz​=9

So,

v⃗(2)=2i^+6j^+9k^\vec v(2) = 2\hat i + 6\hat j + 9\hat kv(2)=2i^+6j^​+9k^
  1. Since the options mention angle from the xxx-axis or yyy-axis, we compare using the relevant components in the xyxyxy-plane.

The projection of velocity on the xyxyxy-plane is

v⃗xy=2i^+6j^\vec v_{xy} = 2\hat i + 6\hat jvxy​=2i^+6j^​
  1. Angle with the yyy-axis:
tan⁡θ=component along xcomponent along y=26=13\tan\theta = \frac{\text{component along }x}{\text{component along }y} = \frac{2}{6} = \frac13tanθ=component along ycomponent along x​=62​=31​

Thus,

θ=tan⁡−1(13)\theta = \tan^{-1}\left(\frac13\right)θ=tan−1(31​)
  1. Angle with the xxx-axis:
tan⁡ϕ=component along ycomponent along x=62=3\tan\phi = \frac{\text{component along }y}{\text{component along }x} = \frac{6}{2} = 3tanϕ=component along xcomponent along y​=26​=3

Thus,

ϕ=tan⁡−1(3)\phi = \tan^{-1}(3)ϕ=tan−1(3)
  1. None of the given options matches tan⁡−1(1/3)\tan^{-1}(1/3)tan−1(1/3) from the yyy-axis or tan⁡−1(3)\tan^{-1}(3)tan−1(3) from the xxx-axis.

If instead the question intended the angle of the velocity vector with the yyy-axis in 3D, then

tan⁡θ=vx2+vz2vy=22+926=856\tan\theta = \frac{\sqrt{v_x^2+v_z^2}}{v_y} = \frac{\sqrt{2^2+9^2}}{6} = \frac{\sqrt{85}}{6}tanθ=vy​vx2​+vz2​​​=622+92​​=685​​

So,

θ=tan⁡−1(856)\theta = \tan^{-1}\left(\frac{\sqrt{85}}{6}\right)θ=tan−1(685​​)

which matches option A.

  1. Therefore, interpreting “direction from y-axis” in 3D, the correct option is:
tan⁡−1(856)\boxed{\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)}tan−1(685​​)​

from the yyy-axis.

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