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?
1 Answer
Solution :-
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Speed of the cyclist = 27 km/h = 27 x (\( \dfrac{5}{18}\)) = 7.5 m/s
Radius of the road = 80 m
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
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