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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8 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 +

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physics ncert solutions class 11th 2023

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