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Motion in A Plane question

2021 · 20 Jul · Shift 1 · Q51
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  5. /2021 · 20 Jul · Shift 1 · Q51

Motion in A Plane question

2021 · 20 Jul · Shift 1 · Q51

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A butterfly is flying with a velocity 424\sqrt 242​ 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 :
  1. A
    12212\sqrt 2122​ m
  2. B
    20 m
  3. C
    3 m
  4. D
    15 m
View written solutionFree

Correct answer: D

  1. Resolve the butterfly's velocity into components

The butterfly flies with speed 42 m/s4\sqrt{2}\ \text{m/s}42​ m/s in the North-East direction.

North-East means equal components toward North and East, so:

vbx=4 m/s(East)v_{bx} = 4\ \text{m/s} \quad (\text{East})vbx​=4 m/s(East) vby=4 m/s(North)v_{by} = 4\ \text{m/s} \quad (\text{North})vby​=4 m/s(North)

because

422=4\frac{4\sqrt{2}}{\sqrt{2}} = 42​42​​=4

So butterfly's velocity vector is:

v⃗b=4i^+4j^\vec v_b = 4\hat i + 4\hat jvb​=4i^+4j^​

  1. Write the wind velocity

Wind is blowing at 1 m/s1\ \text{m/s}1 m/s from North to South.

So the wind velocity is toward South:

v⃗w=−1j^\vec v_w = -1\hat jvw​=−1j^​

  1. Find the resultant velocity of the butterfly

Assuming the butterfly's ground velocity is the vector sum of its own velocity and wind velocity:

v⃗=v⃗b+v⃗w\vec v = \vec v_b + \vec v_wv=vb​+vw​ v⃗=(4i^+4j^)+(−1j^)\vec v = (4\hat i + 4\hat j) + (-1\hat j)v=(4i^+4j^​)+(−1j^​) v⃗=4i^+3j^\vec v = 4\hat i + 3\hat jv=4i^+3j^​

Its magnitude is:

∣v⃗∣=42+32=16+9=5 m/s|\vec v| = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5\ \text{m/s}∣v∣=42+32​=16+9​=5 m/s

  1. Find the displacement in 3 seconds

Since the velocity is constant, displacement magnitude in 333 s is:

s=vt=5×3=15 ms = vt = 5 \times 3 = 15\ \text{m}s=vt=5×3=15 m

  1. Match with the options

15 m15\ \text{m}15 m

So the correct option is:

D: 151515 m

  1. Comparison with stored correct answer

Stored correct answer is D, which matches the derived answer.

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