Question #abf84

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Question #abf84

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Answer 1 · 2016-10-22T19:05:10

${24.6}^{\circ}$ $24.6^@$ , 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
$tan θ = \frac{{(\frac{108 \times 10^{3}}{3600})}^{2}}{200 \times 9.81} = \frac{{(30)}^{2}}{200 \times 9.81}$ $tan theta=((108xx10^3)/3600)^2/(200xx9.81)=(30)^2/(200xx9.81)$
$\Rightarrow tan θ = 0.4587$ $=>tan theta=0.4587$
$\Rightarrow θ = {tan}^{- 1} 0.4587$ $=>theta=tan^-1 0.4587$
$\Rightarrow θ = {24.6}^{\circ}$ $=>theta=24.6^@$ , rounded to one decimal place.

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Question #abf84

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