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4-1.Newton's Laws of Motion
medium
A motor cyclist moving with a velocity of $72\, km/hour$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20$ meters. The acceleration due to gravity is $10 m/sec^2$. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than
A$\theta = {\tan ^{ - 1}}6$
B$\theta = {\tan ^{ - 1}}2$
C$\theta = {\tan ^{ - 1}}25.92$
D$\theta = {\tan ^{ - 1}}4$
Solution
(b)$v = 72\,km/hour = 20\,m/\sec $
$\theta = {\tan ^{ – 1}}\left( {\frac{{{v^2}}}{{rg}}} \right) = {\tan ^{ – 1}}\left( {\frac{{20 \times 20}}{{20 \times 10}}} \right) = {\tan ^{ – 1}}(2)$
$\theta = {\tan ^{ – 1}}\left( {\frac{{{v^2}}}{{rg}}} \right) = {\tan ^{ – 1}}\left( {\frac{{20 \times 20}}{{20 \times 10}}} \right) = {\tan ^{ – 1}}(2)$
Standard 11
Physics
