SOLUTION: The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car traveling 50 mph can stop in 110 ​ft, how many feet will

Algebra ->  Radicals -> SOLUTION: The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car traveling 50 mph can stop in 110 ​ft, how many feet will       Log On


   



Question 1097414: The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car traveling 50 mph can stop in 110 ​ft, how many feet will it take the same car to stop when it is traveling 100 ​mph? It will take how many feet? Please give exact answer!!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
When a variable (like braking distance, d, in feet)
varies directly with some other variable (like the square of the speed, with r=speed measured in mph),
as one variable changes by a factor, so does the other.
As the speed doubles, from 50 mph to 110 mph,
the square of the speed quadruples.
It changes from 50%5E2=2500 to 100%5E2=10000 ,
but we do not need to calculate to know
that when one number doubles its square quadruples.
The square of the speed quadrupled,
and so does the breaking distance,
from 110ft to 4%2A%28110ft%29=highlight%28440ft%29 .
Easy mental math.

If you were asked to write it as a function,
you would define k as a positive constant,
say that
d=k%2Ar%5E2 ,
and substituting 110 for the distance and 50 for the speed,
you would find k :
110=k%2A50%5E2 --> 110=k%2A2500 --> 110%2F2500=k --> k=0.044 .
Then, the function,would be
d=0.044%2Ar%5E2 .
Then, you could use that function,
substituting 100 for r ,
to calculate the answer as
d=0.044%2A100%5E2=0.044%2A10000=440 .
That would require a lot more writing and calculating.