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?

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    Vishal Baghel | Contributor-Level 10

    4 months ago

    4.31

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

    Radius of the road = 80m

    The net acceleration is due to braking and the centripetal acceleration

    Acceleration due to braking = 0.5 m/s2

    Centripetal acceleration a = v2r = (7.5)2/80= 0.703 m/s2

    The resultant acceleration is given by a = sqrt ( at2 + ac2 ) = sqrt ( 0.52 + 0.72 ) = 0.86 m/s2

    tan β = AC / at = 0.7/0.5,  β = 54.5 °

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