Question

15. The road in the question 14 above is

constructed as per the requirements. The

coefficient of static friction between the

tyres of a vehicle on this road is 0.8, will

there be any lower speed limit? By how

much can the upper speed limit exceed in

this case?​

Answer 1

Ans : 88kmph

No upper limit as the road is banked for theta> 45⁰

Answer 2

$\underline{ \underline{ \underline{\textsf{ \textbf{ \red{\qquad Given : \qquad }}}}}}$

• Coefficient of static friction (μ) = 0.8
• Radius of curvature (r) = 72 m
• Acceleration due to gravity (g) = 10 m/s²
• tan θ = 5

$\underline{ \underline{ \underline{\textsf{ \textbf{ \purple{\qquad To Find : \qquad }}}}}}$

• By how much can the upper speed limit exceed in this case ?

$\underline{ \underline{ \underline{\textsf{ \textbf{ \gray{\qquad Solution : \qquad }}}}}}$

$:\implies\sf V_{(minimum)} = \sqrt{rg \Bigg\lgroup\dfrac{\tan( \theta) - \mu }{1 + \mu\tan( \theta)} \Bigg \rgroup}$

$:\implies\sf V_{(minimum)} = \sqrt{72 \times 10 \Bigg\lgroup\dfrac{5 - 0.8 }{1 +0.8 \times 5} \Bigg \rgroup}$

$:\implies\sf V_{(minimum)} = \sqrt{720 \Bigg\lgroup\dfrac{4.2 }{1 +4} \Bigg \rgroup}$

$:\implies\sf V_{(minimum)} = \sqrt{\dfrac{3024}{5} }$

$:\implies\sf V_{(minimum)} = \sqrt{604.8 }$

$:\implies\sf V_{(minimum)} = 24.592 \:m/s$

$:\implies\sf V_{(minimum)} = \dfrac{24.592 \times 18}{5} \:km/hr$

$:\implies\sf V_{(minimum)} = 88.5312\:km/hr$

$:\implies \underline{ \boxed{\textsf{\textbf{ V _{(\text {minimum})} \approx 88 km/hr}}}}$

As θ ≽ 45° there is no upper limit exceed in the case.