A mosquito is moving with a velocity v⃗ = 0.5t²î + 3tĵ + 9k̂ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?

Solution

Physics sir · 4 steps

Put t = 2 into the velocity first, then compare the part along the y-axis with the part perpendicular to it.

1. Velocity at 2 s. Put t = 2 into each component: vₓ = 0.5 × 4 = 2, v_y = 3 × 2 = 6, v_z = 9. So v⃗ = 2î + 6ĵ + 9k̂ m s⁻¹, with magnitude √(4 + 36 + 81) = 11 m s⁻¹.
2. Angle with an axis. In 3D, the angle θ with an axis has tan θ equal to the part of v⃗ perpendicular to that axis, divided by the part along it. For the y-axis, the perpendicular part combines vₓ and v_z.

   tan θ_y = √(vₓ² + v_z²)/v_y = √(4 + 81)/6 = √85/6

3. Check. With cosines, cos θ_y = 6/11, so sin θ_y = √(121 - 36)/11 = √85/11. That gives the same tan θ_y = √85/6.
4. So. The mosquito moves at tan⁻¹(√85/6) from the y-axis, which is option (a).

The idea to keep. For the angle with one axis in 3D, use tan = (root of the sum of squares of the other two components) / (component along that axis). Or find cos = component / magnitude.

Answer from the NTA final answer key. Solution by PrepWiser.