Question

A racer can bend his bike at a maximum angle of ${53}^{\circ }$ with the vertical plane. If he approaches a curve driving with a velocity of $126\text{km/h}$, what should be the turning radius to make a safe turn?

• A. $100.81\text{m}$
• B. $156.25\text{m}$
• C. $90\text{m}$
• D. $94.23\text{m}$

Answer 1

The correct option is D $94.23\text{m}$

$v=126\text{km/h}=35\text{m/s}$
For a safe turn without skidding,
$\tan \theta =\frac{v^2}{rg}$
$\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{v^2}{rg}$
$\Rightarrow \frac{\sin {53}^{\circ }}{\cos {53}^{\circ }}=\frac{v^2}{rg}$
$\Rightarrow \frac{\frac{4}{5}}{\frac{3}{5}}=\frac{{35}^2}{r\times 10}$
$\Rightarrow r=\frac{1225}{1.3\times 10}=94.23\text{m}$