A car is travelling with liner velcoity v on a circular road of radius R. If its speed is increasing at the rate of a `ms^(-1)` , then the net acceleration will be.

Options

A. `(v^(2))/(R ) +a`
B. `(v^(2))/(R ) - a`
C. `sqrt(((v^(2))/(R))^(2)+a)`
D. `sqrt(((v^(2))/(R))^(2)+a^(2))`

Answer

D

Explanation

Speed in increasing at the rate of a `ms^(-2)` . So, linear acceleration =a
and tangential acceleration ` = v^(2)//R`
Hence, net acceleration `= sqrt(((v^(2))/(R))^(2) +a^(2))`