SOLUTION: The stopping distance s of a car varies directly as the square of its speed v. If a car traveling at 30 mph requires 45 ft to stop, find the stopping distance for a car traveling a

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Question 1204790: The stopping distance s of a car varies directly as the square of its speed v. If a car traveling at 30 mph requires 45 ft to stop, find the stopping distance for a car traveling at 45 mph.

Found 3 solutions by mananth, greenestamps, josgarithmetic:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
The stopping distance s of a car varies directly as the square of its speed v. If a car traveling at 30 mph requires 45 ft to stop, find the stopping distance for a car traveling at 45 mph.

d= kv^2
car traveling at 30 mph requires 45 ft to stop,
45= k*(30)^2
k= 45/900= 1/20
stopping distance for a car traveling at 45 mph.
d=k*v^2
d = (1/20) * 45^2
d=101.25 ft

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

The ratio of speeds is 45:30 = 3:2.

Since the stopping distance varies as the square of the speed, the ratio of stopping distances is 3^2:2^2 = 9:4.

ANSWER: 45*(9/4) = 405/4 = 101.25 feet

Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
This or something like it was solved about a couple days ago.

s, stopping distance
v, speed of car
k, variation constant

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