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Proportion and velocity

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by manelgirona » Mon Feb 15, 2010 3:09 pm
If the distance required for an automobile to stop varies directly as the square of its speed, what is the distance required for an automobile driven at 60 kilometers an hour to stop if it takes 10 meters to stop the automobile at 20 kilometers an hour?

a) 30 meters
b) 36 meters
c) 60 meters
d) 90 meters
e) 120 meters
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by djkvakin » Mon Feb 15, 2010 6:42 pm
the square of the 20mph speed is 400 and we know that the breaking distance is 10. from here we set up a proportion:
10/400=x/3600. Solve for x
we get x=90 = D
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by Stuart@KaplanGMAT » Mon Feb 15, 2010 6:50 pm
manelgirona wrote:If the distance required for an automobile to stop varies directly as the square of its speed, what is the distance required for an automobile driven at 60 kilometers an hour to stop if it takes 10 meters to stop the automobile at 20 kilometers an hour?

a) 30 meters
b) 36 meters
c) 60 meters
d) 90 meters
e) 120 meters
Triple the speed means 3^2 times the stopping distance.

9 * 10 = 90 metres!
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by shashank.ism » Mon Feb 15, 2010 9:25 pm
manelgirona wrote:If the distance required for an automobile to stop varies directly as the square of its speed, what is the distance required for an automobile driven at 60 kilometers an hour to stop if it takes 10 meters to stop the automobile at 20 kilometers an hour?

a) 30 meters
b) 36 meters
c) 60 meters
d) 90 meters
e) 120 meters
D = function(v^2)
so D = k V^2
10 = k (20^2)

Hence[spoiler] D= 60^2 x 10 /20^2 = 90 meters Ans D[/spoiler]
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