Motion in a Plane

Question : 31

Total: 32

A cyclist is riding with a speed of 27‌km‌h–1. 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 0.5ms–1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Solution:

Here, v=27‌km‌h−1=7.5ms−1 ; r=80m
Centripetal acceleration, ac=

(7.5)2

=0.7ms−2
Suppose that the cyclist applies brakes at the point A of the circular turn. Then, retardation produced dueto the brakes, say aT will act opposite to the velocity, v figure.
Thus, aT=–0.5ms–2
Therefore, total acceleration is given by
a=√ac2+aT2 =√(0.7)2+(−0.5)2 =√0.49+0.25=√0.74=0.86ms−2
Ifθ is the angle between the total acceleration and the velocity of the cyclist, then tan‌θ=

0.7

0.5

=1.4 or θ=54°28′

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