A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of a radius of 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?

Asked by Abhisek | 1 year ago |  196

##### Solution :-

Speed of the cyclist = 27 km/h = 27 x ($$\dfrac{5}{18}$$) = 7.5 m/s

The net acceleration is due to the braking and the centripetal acceleration

Due to braking, a =  0.50 m/s2

Centripetal acceleration, a = $$\dfrac{v^2}{2}$$ = $$\dfrac{7.5^2}{80}$$ = 0.70 m/s2

since the angle between ac and aT is 90o, the resultant acceleration is given by

$$a = \sqrt{a_{t}^{2}+a_{c}^{2}}= \sqrt{(0.5)^{2}+(0.7)^{2}}$$

= 0.86 m/s2

and tan β = $$\dfrac{a_c}{a_t}$$ = $$\dfrac{0.7}{0.5}$$= 1.4

β = tan -1 (1.4) = 54.50 from the direction of velocity

Answered by Pragya Singh | 1 year ago

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