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

Here, ` v= 27 km h^(-10 = 27 xx 9 1000 m) xx ( 60 xx 60 s^(-10 = ms^(-20`
Centripetal acceleration ` a_c = v^2 /r (( 7.5)^2)/(80) = 0.7 ms^(-20` ltbRgt Let the cyclist applies the brakes at the point (P) of the circular turn, then tangential acceleration ` a_T` ( which will be begative) will act opposite to velcoity, Fig. 2 (NCT) . 25.
.
Acceleration along the tangent, ` a_T=0.5 ms^(-2)`
Angle between both the acceleration is ` 90^@`
Therefore, the magnitude of resulatnt accelerattion, ` a= sqrt ( a_c^2 = a_T^2) = sqrt (9o. 7) ^@ + ( 0.5)^2) = 0.86 ms^(-2)`
Let the resultant acceleration maken an angle ` beta` with the tangent i.e. the direction of net acceleration of the
cyclist, then , ` tan beta = a_V /a_T = (0.7)/(0.5) = 14` or ` beta = 54^@ 28' `.