A cyclist is riding with a speed of `27 km h^(-1)`. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate `0.5 ms^(-1)`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Options

A. `0.86 m//s^(2)`
B. `0.43 m//s^(2)`
C. `1.24 m//s^(2)`
D. `1.76 m//s^(2)`

Answer

A

Explanation

`v=27xx(5)/(18)=(45)/(6)=(15)/(2)`
` a=sqrt(a^(2)+((v^(2))/(r))^(2))=sqrt(0.25+(15xx15)/(4xx80))`
`=sqrt(0.25+(45)/(64))`
`=sqrt(0.25+0.49)=sqrt(0.25+0.70)=sqrt(0.95)`
`0.86m//s^(2)`