Question
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
Speed of the cyclist, v = 27 km/h = 7.5 m/s
Radius of the circular turn, r = 80 m
Centripetal acceleration is,
The question can be illustrated in the figure as shown below:
Consider, 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/s2.
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.
The angle between ac and aT is 900. Therefore, the resultant acceleration a is given by,
