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If the maximum allowable speed of a vehicle is 20 m/s and can accelerate at 0.25 m/s2, then what must be the radius of the circular curve and length of transition curve such that ratio of centrifugal force on vehicle to its weight is 1 ∶ 4?

(Take g = 10 m/s2)


1. 250 m, 320 m
2. 250 m, 160 m
3. 160 m, 500 m
4. 160 m, 92 m

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Best answer
Correct Answer - Option 4 : 160 m, 92 m

Concept:

The length of the transition curve on the basis of rate of change of acceleration is given as :

\(L = \frac{{{V^3}}}{{RC}}\)

Where

V is the Design Speed (m/sec)

R is the radius of transition curve at point where it meets to circular curve (m)

C is the jerk or rate of change of acceleration. (m/sec3)

Further,

The ratio of centrifugal force (P) and weight of vehicle (W) , which is called impact factor, is given as:

\(\frac{P}{W} = \;\frac{{{V^2}}}{{gR}}\)

Calculation:

Given:

P/W = 1/4

g = 10 m/s2, V = 20 m/s = 72 kmph

Acceleration a = 0.25 m/s2

Radte of change of centrifugal Acceleration (C) = 80/(75 + V) 

where V is in kmph

i.e.  (C) = 80/(75 + V) = 80/(75 + 72) = 0.544 m/sec3

We know that \(\frac{P}{w} = \frac{{{V^2}}}{{{g_R}}}\) 

\(\frac{1}{4} = \frac{{{{\left( {20} \right)}^2}}}{{10\; \times\; R}} \Rightarrow R = 160\;m\)

Also,

\(L = \frac{{{V^3}}}{{RC}} = \frac{{{{\left( {20} \right)}^3}}}{{160\; \times \;0.54}} = 91.91\;m\)

∴ R = 160 m & L = 92 m is most appropriate answer.

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