An unbanked circular highway curve on the level ground makes a turn of 90° . The highway carries traffic at 108 km h- 1 , and the centripetal force on a vehicle is not to exceed (1/1 0) of its weight. What is the approximate minimum length of the curve, in km?

Question

An unbanked circular highway curve on the level ground makes a turn of 90° . The highway carries traffic at 108 km h- 1 , and the centripetal force on a vehicle is not to exceed (1/1 0) of its weight. What is the approximate minimum length of the curve, in km? (A) 1.41km (B) 1km (C) 0.6km (D) None of these

Tardigrade Admin · Accepted Answer

(m v2/R)=(m g/1 0) R=(1 0 v2/g)=( text108 × (5/1 8))2=900 textm , Length (π R/2)= text1.413 km

Q. An unbanked circular highway curve on the level ground makes a turn of $9 0^{\circ}$ . The highway carries traffic at $108 km h^{- 1}$ , and the centripetal force on a vehicle is not to exceed $\frac{1}{10}$ of its weight. What is the approximate minimum length of the curve, in km?

$1.41 km$

$1 km$

$0.6 km$

None of these

$\frac{m v ^{2}}{R} = \frac{m g}{10}$
$R = \frac{10 v ^{2}}{g} = (108 \times \frac{5}{18})^{2} = 900 m$ ,
Length $\frac{π R}{2} = 1.413 km$

Q. An unbanked circular highway curve on the level ground makes a turn of 90∘ . The highway carries traffic at 108kmh−1 , and the centripetal force on a vehicle is not to exceed 101​ of its weight. What is the approximate minimum length of the curve, in km?

1.41km

1km

0.6km