StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A butterfly flies at 4*sqrt(2) m/s in the north-east direction relative to the wind. The wind blows at 1 m/s from north to south (i.e., in the south direction). Find the magnitude of the resultant displacement of the butterfly after 3 seconds.

  1. 3 m
  2. 20 m
  3. 12*sqrt(2) m
  4. 15 m

Correct answer: 15 m

Solution

Velocity of butterfly relative to ground: East component = 4*sqrt(2) * cos(45) = 4 m/s. North component = 4*sqrt(2) * sin(45) - 1 = 4 - 1 = 3 m/s (subtracting wind going south). Speed = sqrt(4² + 3²) = sqrt(16+9) = sqrt(25) = 5 m/s. Displacement in 3 s = 5 * 3 = 15 m.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →