Question: A car travelling at 130kph locks up its wheel and skids up a 3% incline before crashing into a st...
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Velocity of car = 130 km/hr = (5/18)×130 mts/sec
= 36.11 m/s
Upgradient n = 3% = 0.03
Skid Marks leading to pedestal = 90 mts
Coefficient of friction f = 0.35 (a)
If the car had not hit the pedestal
Distance of skid = (V^2)/(2g(f-n)) = ((36.11)^2)/(2*9.81(0.35-0.03)) = 207.7 metres
total distance of skid = 207.7 mts
It would have further skidded a distance of 207.7 - 90 = 117.7 mts (b)
speed at impact
from the equation V^2 - U^2 = 2g(f-n)S
Initial velocity = 36.11 m/s
Distance S = 90 meters
f= 0.35 n= 0.03
(36.11)^2 - U^2 = 2×9.81×(0.35-0.03)×90
U^2 = 1304 - 565= 739
U = 27.2 m/s
Hence velocity of car at impact = 27.2 m/s = 97.9 km/hr
(c)
Actual stopping sight distance = braking distance = 207.7 mts
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Question: A vertical summit curve has tangent grades of +2.8% and -1.6%. The curve will be designed such th...
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Question: A vertical curve must begin at the center of manhole 1 (sta. 56+000, at elev.60m ) and end at man...