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^(-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)=0.5 ms^(-2)`
`a=sqrt(2_(C)^(2)+a_(T)^(2))=0.86 ms^(-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'`