Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Answers (1)

R=2.5 m                   a=15  m/s2

From figure, centripetal acceleration  

a_c =a\cos30= 15 \times 0.866 = 13 m/s^2 

Using  a_c = \frac{v^2}{R} 

\therefore 13 = \frac{v^2}{2.5} \\ \\ \implies v= 5.7 m/s

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question