A car of mass m moves in a horizontal path of radius r metre .At instant its speed is v m/s and is increasing at the end rate of a `m//sec^(2)` .Then the acceleration of the car is .

Options

A. `(v^(2))/t`
B. a
C. `sqrt(a^(2)+(v^(2)/r)^(2))`
D. `sqrt(a+(v^(2))/r)`

Answer

B

Explanation

Radial acceleration `a_(r ) = (v^(2))/r`
Tangential acceleration `a_(r )=a`
Resultant acceleration = a= `sqrt(a_(r )^(2)+a_(r )^(2)+2a_(r )a_(t) cos theta)`
But `theta = 90^(@)`
` :. cos theta = cos 90^(@) = 0 and a=sqrt(a_(r)^(2)+a_(r)^(2))=sqrt(((v^(2))/r)^(2)+a^(2))` .