The distance travelled by a vehicle at speed V on an upgrade after the brake is applied is equal to the distance travelled by the same vehicle at 0.75 V on a downgrade of the same amount. The grade will be :
(Take f=0.30)
A
8.4%
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B
6.8%
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C
4.8%
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D
9.6%
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Solution
The correct option is A 8.4% Braking distance =V22g(f±n)
From equation, V22g(f+n)=(0.75V)22g(f−n) f+nf−n=169 f+n+(f−n)f+n−(f−n)=16+916−9[Bycomponendo−dividendo] fn=257 n=725×0.3=0.084=8.4%