Question

A road is $7.5\text{m}$ wide and its outer edge is raised by $1.5\text{m}$ over the inner edge. The radius of curvature is $50\text{m}$. If $g=10{\text{m/s}}^2$, for what speed of the car is this road suited?

• A. $2.5\text{m/s}$
• B. $5\text{m/s}$
• C. $10\text{m/s}$
• D. $15\text{m/s}$

Answer 1

The correct option is C $10\text{m/s}$

By the FBD of the car,
By balancing the vertical forces we get,
$N\cos \theta =mg...\left(i\right)$ and here the horizontal component of the normal provides the necessary centripetal force.
$N\sin \theta =\frac{m{v}^2}{r}...\left(ii\right)$, by dividing both the above equation $\left(i\right)$ and $\left(ii\right)$ we get,
$\tan \theta =\frac{v^2}{rg}...\left(iii\right)$

For finding the angle $\theta$,It is given in the problem that $L=7.5\text{m}$ and $h=1.5$.
Here we get $\sin \theta =\frac{h}{L}=\frac{1.5}{7.5}=\frac{1}{5}$, for small angle $\theta$ we know that,
$\sin \theta =\tan \theta =\theta$, by substituting this value in equation $\left(iii\right)$ we get,
$\frac{1}{5}=\frac{v^2}{rg}\Rightarrow v^2=\sqrt{\frac{rg}{5}}=\sqrt{\frac{50\times 10}{5}}=10\text{m/s}$

Hence option C is the correct answer