A car is moving along a circular path of radius 500 m with a speed of `30 ms^(-1)` .If at some instant , its speed increases at the rate of `2 ms^(-2)` , then at that instant the magnitude of resultant acceleration will be

Options

A. `4.7 ms^(-2)`
B. `3.8 ms^(-2)`
C. `3 ms^(-2)`
D. `2.7 ms^(-2)`

Answer

D

Explanation

Centripetal acceleration
`a_(c )=(v^(2))/r =((30)^(2))/500 =1.8 m//s^(2)`
Tangential acceleration `a_(t) =2ms^(-2)`
` :.` Resultant acceleration `a=sqrt(a_(t)^(2)+a_(c )^(2))=sqrt((1.8)^(2)+2^(2))=2.7 m//s^(2)`