A cyclist is riding with a speed of 36 ,km/h. | vedclass

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(D) Initial speed $V = 36 \,km/h = 36 	imes \frac{5}{18} = 10 \,m/s$. Radial acceleration $a_r = \frac{V^2}{R} = \frac{10^2}{50} = \frac{100}{50} = 2

Vedclass · Accepted Answer

$\frac{\sqrt{17}}{2} \,m/s^2, 	an^{-1}(4)$

Vedclass · Answer

(D) Initial speed $V = 36 \,km/h = 36 	imes \frac{5}{18} = 10 \,m/s$. Radial acceleration $a_r = \frac{V^2}{R} = \frac{10^2}{50} = \frac{100}{50} = 2 \,m/s^2$. Tangential acceleration $a_t = 0.5 \,m/s^2$. Net acceleration $a_{net} = \sqrt{a_r^2 + a_t^2} = \sqrt{2^2 + 0.5^2} = \sqrt{4 + 0.25} = \sqrt{4.25} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \,m/s^2$. Direction $	heta$ with respect to the radial vector is given by $	an 	heta = \frac{a_t}{a_r} = \frac{0.5}{2} = \frac{1}{4}$. However, the angle with respect to the tangential vector is $	an \phi = \frac{a_r}{a_t} = \frac{2}{0.5} = 4$, so $\phi = 	an^{-1}(4)$.

$A$ cyclist is riding with a speed of $36 \,km/h$. As he approaches a circular turn on the road of radius $50 \,m$, he applies brakes and reduces his speed at the constant rate of $0.5 \,m/s^2$. The magnitude and direction of the net acceleration of the cyclist on the circular turn are respectively:

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