1.

If the velocity of a car in increased by 20%, then the minimum distance in which it can be stopped increases by:

Answer»

0.44
0.55
0.66
0.88

Solution :Here`0^(2)-v^(2)= -2aS` or `v^(2)PROPS`
or `S/S=(1+(20)/(100))^(2)=(36)/(25)`
or `((S)/(S)-1)xx100=((36)/(25)-1)xx100`=44%


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