Question #e5a6d

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Question #e5a6d

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Answer 1 · 2016-10-24T22:30:08

${22.4}^{\circ}$ $22.4^@$ , rounded to one decimal place.

Explanation:

If banking angle is $θ$ $theta$ then we have the following expression connecting various variables
$tan θ = \frac{v^{2}}{r g}$ $tan theta=v^2/(rg)$
where $v$ $v$ is the velocity of the vehicle, $r$ $r$ is the radius of curvature and $g$ $g$ is the acceleration due to gravity equal to $9.81 m s^{- 2}$ $9.81ms^-2$ .

Inserting given values after converting speed to SI units, we get
$tan θ = \frac{{(\frac{45 \times 1609.344}{3600})}^{2}}{100 \times 9.81} = 0.4125$ $tan theta=((45xx1609.344)/3600)^2/(100xx9.81)=0.4125$
$\Rightarrow θ = {tan}^{- 1} 0.4125$ $=>theta=tan^-1 0.4125$
$\Rightarrow θ = {22.4}^{\circ}$ $=>theta=22.4^@$ , rounded to one decimal place.

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