NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A point object moves on a circular path such that distance covered by it is given by function S = t 2 2 + 2 t meter ( t in second). The ratio of the magnitude of acceleration at t = 2 s and t = 5 s . is 1 : 2 then the radius of the circle is
Options
- A1 m
- B3 51 m
- C51 m
- D3 m
Correct answer
B. 3 51 m
Step-by-step solution
a t = 1 m / s 2 v = t + 2 at t = 2 s a 2 = 1 2 + ( 2 + 2 ) 2 R 2 a 5 = 1 2 + ( 5 + 2 ) 2 R 2 2 1 + 1 6 2 R 2 = 1 + 4 9 2 R 2 4 + 4 1 6 2 R 2 = 1 + 4 9 2 R 2 3 = 4 9 2 R 2 - 4 1 6 2 R 2 ⇒ R = 3 51 m