A cylclist 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 of `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Explanation

This is a case of a body moving in a circular path with variable speed. So it has centripetal acceleration `a_( c)` and tangential `a_(T)` which are perpendicular to each other. The resultant acceleration.

`a = sqrt(a_( c)^(2) + a_(T)^(2)), a_( c) = v^(2)/r`
`v = 27 km//h = (27 xx 1000)/(60 xx 60) = 7.5 ms^(-1), r= 80 m`
`a_( c) = (7.5)^(2)/80 = 0.703 ms^(-2)`
Tangential acceleration `a_(T) = (Deltav)/(Deltat) = 0.5/1 = 0.5 ms^(-2)`
`a = sqrt(a_(c )^(2) + a_(T)^(2)) = sqrt(0.703^(2) + 0.5^(2)) = 0.8627 ms^(-2)`
Let the resultant acceleration make an angle a with the radius. Then
`tan alpha = a_(T)/a_( c) = 0.5/(0.703) = 0.7112, alpha = tan^(-1) (0.7112) = 35^(@)30.`.