NTA Abhyas JEE Main2020PhysicsMotion in Two DimensionsPractice
The position vector of the particle is r t = a cos ω t i ^ + a sin ω t j ^ , where a and ω are real constants of suitable dimensions. The acceleration is
Options
- APerpendicular to the velocity
- BParallel to the velocity
- CDirected away from the origin
- DPerpendicular to the position vector
Correct answer
A. Perpendicular to the velocity
Step-by-step solution
Given that, r t = a cos ω t i ^ + a sin ω t j ^ ∵ v → = d r t d t = - a ω sin ω t i ^ + a ω cos ω t i ^ j a → = d v → d t = - a ω 2 cos ω t i ^ - a ω 2 cos ω t j ^ a → . v → = a 2 ω 3 sin ω t cos ω t - a 2 ω 3 sin ω t cos ω t ⇒ a → . v → = 0 Above result implies that acceleration is perpendicular to velocity.