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Asked by Jyoti Last Modified
Answer 0.86 m/s2; 54.46° with the direction of velocity
Speed of the cyclist, 
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
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/s2.
This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between 
is 90°, the resultant acceleration a is given by:
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