A cyclist is riding with a speed of `27kmh^(-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 `0.5ms^(-2)` . What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Explanation

`a_(c)=(v^(2))/(r)=0.7ms^(-2)`
`a_(T)=0.5ms^(-2)`
`a=sqrt(a^(2)-a^(2))=0.86ms^(-2)`
If `theta` is the angle between the net acceleration and the velocity of the cyclist, then
`theta=tan^(-1)((a_(c))/(a_(T)))=tan^(-1)(1.4)=54^(@)28.`