Question:
Angle of banking for a vehicle speed of $ 10\,m/s $ for a radius of curvature $ 10\,m $ is (assume $ g=10\,m/s^{2} $ )
Angle of banking for a vehicle speed of $ 10\,m/s $ for a radius of curvature $ 10\,m $ is (assume $ g=10\,m/s^{2} $ )
Updated On: Jun 14, 2022
- $ 30^{\circ} $
- $ \tan^{-1}\left( \frac{1}{2} \right) $
- $ 60^{\circ} $
- $ 45^{\circ} $
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The Correct Option is D
Solution and Explanation
Angle of banking is given by $\tan \theta=\frac{v^{2}}{r g}$
where $v$ is velocity, $r$ the radius of circular path and $g$ acceleration due to gravity. Given,
$v=10 \,m / s , r=10 \,m , g=10 \,m / s ^{2} $
$\therefore \tan \theta=\frac{10 \times 10}{10 \times 10}=1 $
$\Rightarrow \theta=45^{\circ}$
where $v$ is velocity, $r$ the radius of circular path and $g$ acceleration due to gravity. Given,
$v=10 \,m / s , r=10 \,m , g=10 \,m / s ^{2} $
$\therefore \tan \theta=\frac{10 \times 10}{10 \times 10}=1 $
$\Rightarrow \theta=45^{\circ}$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration