A cyclict is riding with a speed of 27 km/h. As he approaches a circular turn on the road of raidius 80m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. 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 m) xx 60 xx 60 s)^(-1) = 7.5 ms^(-1), r = 80 m` centripetal acceleration
`a_(c) = (v_(2))/(r)`
`= ((7.5)^(2))/(80) = 0.7 ms^(-2)`

Let the cyclist applies the brakes at the point P of the circular turn, then tangential acceleration `a_(T)` will act opposite to velocity. Acceleration along the tangent , `a_(T) = 0.5 ms^(-2)` Angle between both the acceleration is `90^(@)` Therefore, the magnitude of 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 acceleration make an angle `beta` with the tangent i.e. the direction of velocity of the cyclist, then,
`tan beta = (a_(c))/(a_(T)) = (0.7)/(0.5) = 1.4`
or `beta = 54^(@)28.`