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 | 197

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/s^{2}

Centripetal acceleration, a = \(
\dfrac{v^2}{2}\) = \(
\dfrac{7.5^2}{80}\) = 0.70 m/s^{2}

since the angle between a_{c} and a_{T} is 90^{o}, the resultant acceleration is given by

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

= 0.86 m/s^{2}

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

β = tan^{ -1} (1.4) = 54.5^{0} from the direction of velocity

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