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

`v=27 km//h=(27xx1000)/(3600)=7.5 m//s`
radius =`r=80m`
centripetal acceleration `a_(c)=(v^(2))/(r)=(7.5xx7.5)/(80)`
`a_(c)=0.7 m//s^(2)`
Let brakes are applied at point p .
`therefore`Acceleration along tangent `=a_(T)=0.5 ms^(-2)`
Angle between ` a_(c) and a_(T)=90^(@)`
The resultant acceleration `=a=sqrt(a_(c)^(2)+a_(T)^(2))`
`=sqrt((0.7)^(2)+(0.5)^(2))`
`=0.86 ms^(-2)`
Let the resultant aceeleration make an nagle `alpha` with the tangent, then
`tan alpha=(a_(c))/(a_(T))=(0.7)/(0.5)=1.4`

`alpha =tan ^(-1)(1.4)`
`=54.46^(@)`