Guest

A point with deacceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in magnitude. at the initial moment t=0, the velocity of the point is V 0 . The velocity of point will be:(S is the distance travelled)

A point with deacceleration along the circle of radius R so that at any moment of time its tangential and normal acceleration are equal in magnitude. at the initial moment t=0, the velocity of the point is V0 . The velocity of point will be:(S is the distance travelled)


 

Grade:11

1 Answers

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
12 years ago

Dear Sanket,

Ans:- The eq of motion is - dv/dt=v²/R

Solving we get,1/v  - 1/vº=t/R ( where vº is the initial velocity)

Or v=vº/(1+vºt/R)

 

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.


All the best Sanket !!!

 


Regards,

Askiitians Experts

SOUMYAJIT IIT_KHARAGPUR

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free