Calculate the optimum speed & the maximum speed with which a car can travel along a curve of radius $80 m$ , If the road is banked at an angle g $21^{o} 48^{1}$ & the friction between the tyres & the road is $0.5$ . $[g = 9.8 m / s^{2}]$ .

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Solution

$tan θ = \frac{v^{2}}{r g}$

$v_{o p t i m u m} = \sqrt{r g tan θ}$

$= \sqrt{80 \times 10 \times 0.38}$

$= 17.43$ m/s

$tan (21^{0} 48^{'}) = 0.38$

$V_{m a x} = \sqrt{\frac{r g (tan θ + μ_{s})}{1 - μ_{s} tan θ}}$

$V_{m a x} = \sqrt{\frac{800 (0.85)}{(1 - 0.19)}}$

$V_{m a x} = 28.96$ m/s

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Calculate the optimum speed & the maximum speed with which a car can travel along a curve of radius 80m, If the road is banked at an angle g 21o481 & the friction between the tyres & the road is 0.5. [g=9.8m/s2].

tanθ=v2rgvoptimum=√rgtanθ =√80×10×0.38 =17.43 m/stan(21048′)=0.38Vmax=√rg(tanθ+μs)1−μstanθVmax=√800(0.85)(1−0.19)Vmax=28.96 m/s

Calculate the optimum speed & the maximum speed with which a car can travel along a curve of radius $80 m$ , If the road is banked at an angle g $21^{o} 48^{1}$ & the friction between the tyres & the road is $0.5$ . $[g = 9.8 m / s^{2}]$ .

$tan θ = \frac{v^{2}}{r g}$

$v_{o p t i m u m} = \sqrt{r g tan θ}$

$= \sqrt{80 \times 10 \times 0.38}$

$= 17.43$ m/s

$tan (21^{0} 48^{'}) = 0.38$

$V_{m a x} = \sqrt{\frac{r g (tan θ + μ_{s})}{1 - μ_{s} tan θ}}$

$V_{m a x} = \sqrt{\frac{800 (0.85)}{(1 - 0.19)}}$

$V_{m a x} = 28.96$ m/s