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Let t be the time passed and x be distance covered by the two ships after which the distance between them becomes shortest. Distance(s) between the ship is given by, S = √(100 − x)² + x². For S to be minimum, dS/dt = 0 ⇒ dS/dt = 1 / 9[(100 − x)² + x²]¹/² ⇒ [−2(100 − x) + 2x] = 0 ⇒ 4x − 200 = 0 ⇒ x = 50 m. So, after both the ships have covered 50 m, distance between them becomes shortest. Time taken for it, will t = x / 10 = 50 / 10 = 5hr.
- Position vector r̄ = cosωt î + sinωt ĵ.
- Velocity vector v̄ = −ωsinωt î + ωcosωt ĵ.
- Acceleration vector ā = −ω² cosωt î − ω² sinωt ĵ = −ω² r̄.
- Nothing actually moves in the direction of the angular velocity vector ω̄. The direction of ω̄ simply represents that the circular motion is taking place in a plane perpendicular to it.
Correct answer: Acceleration vector ā = −ω² cosωt î − ω² sinωt ĵ = −ω² r̄.
Solution
Option C is correct because the acceleration vector in uniform circular motion is always directed towards the center of the circle, and its magnitude is proportional to the square of the angular velocity and the radius. This matches the given expression ā = −ω² r̄.
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