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

Answer: (C)

$\sqrt{ a^2 + \bigg( \frac{ v^2 }{ r } \bigg)^2 }$

Answer 2

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' = $\sqrt{ a_r^2 + a_t^2 } = \sqrt{ a^2 + \bigg( \frac{ v^2 }{ r } \bigg)^2 }$