__Question 4.31__:

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 ?

__Solution__:

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:

^{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}

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