Question

A car of mass m moves in a horizontal circular path of radius $r\,m$. At an instant its speed is v $ ms^{ - 1 } $ and is increasing at a rate of a $ ms^{ - 2}$ . Then the acceleration of the car is

• A. $ \frac{ v^2 }{ r } $
• B. a
• C. $ \sqrt{ a^2 + \bigg( \frac{ v^2 }{ r } \bigg)^2 } $
• D. $ \sqrt{ u + \frac{ v^2 }{ r }} $

Answer 1

The Correct Option is C

Radial acceleration $ a_r = \frac{v^2 }{ r } $ Tangential acceleration $ a_t $ = a $ \therefore $ Resultant acceleration $ a' = \sqrt{ a_r^2 + a_t^2 + 2 a_r a_t \, \cos \, \theta } $ But here $ \theta = 90^\circ $ $ \therefore cos \, \theta = \cos \, 90^\circ = 0 $ and a' = $ \sqrt{ a_r^2 + a_t^2 } = \sqrt{ a^2 + \bigg( \frac{ v^2 }{ r } \bigg)^2 } $