Question

A road is 8 m wide. Its radius of curvature is 40 m. The outer edge is above the lower edge by a distance of 1.2 m. This road is most suited for a velocity of

• A. 5.7 m/s
• B. 7.7 m/s
• C. 36.1 m/s
• D. 9.7 m/s

Answer 1

Answer: (B)

$\tan \theta =\frac{v^2}{rg}$ $\sin \theta =\frac{v^2}{rg}$ $(\sin \theta \approx \tan \theta ,forsmall\theta )$ $\therefore$ $\frac{1.2}{8}=\frac{v^2}{40\times 9.8}$
$v^2=1.2\times 49$ $v=\sqrt{1.2\times 49}$ $=\sqrt{58.8}=7.7m/s$