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^(-1) = 27 xx 1000//(60 xx 60)`
` = 15//2 ms^(-10 , r =80 m`
Centipetal acceleration,
` a-c = v^2/r = ((15)/2)^2 xx 1/(80) = 0.7 ms^(-2)`
Tangential accelertion,
:. Effective accelerrtion,
` a= sqrt( a_c^2 _a_T^2) = sqrt ((0.7)^2 + (0.5)^@)` .