### Q.31:- A cyclist is riding with a speed of 27 km/h. 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 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

**Answer:-**

0.86 m/s^{2}; 54.46° with the direction of velocity

Speed of the cyclist, v = 27 km/h = 7.5 m/s

Radius of the circular turn, *r* = 80 m

Centripetal acceleration is given as:

a_{c} = v^{2} / r

= (7.5)^{2} / 80 = 0.7 ms^{-2}

The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s

^{2}.This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.

Since the angle between a_{c} and a_{T} is 90^{0}, the resultant acceleration *a* is given by:

a = (a_{c}^{2} + a_{T}^{2})^{1/2}

= ( (0.7)^{2} + (0.5)^{2} )^{1/2}

= (0.74)^{1/2} = 0.86 ms^{-2}

tan θ = a_{c} / a_{T}

where θ is the angle of the resultant with the direction of velocity.

tan θ = 0.7 / 0.5 = 1.4

θ = tan^{-1} (1.4) = 54.56^{0}