SOLUTION: The braking distance (in feet) of a car going V mph is given by {{{d(v)=v^2/20+v}}} v is greater or equal to 0. how fast would the car have been traveling for a braking distance

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Question 332220: The braking distance (in feet) of a car going V mph is given by d%28v%29=v%5E2%2F20%2Bv v is greater or equal to 0.
how fast would the car have been traveling for a braking distance of 150feet? round to nearest mile per hour.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The braking distance (in feet) of a car going V mph is given by
d%28v%29=v%5E2%2F20%2Bv v is greater or equal to 0.
how fast would the car have been traveling for a braking distance of 150feet?
round to nearest mile per hour.
:
Write it:
v%5E2%2F20%2Bv = 150
multiply by 20, results
v^2 + 20v = 20(150)
:
v^2 + 20v = 3000
:
v^2 + 20v - 3000 = 0
Solve for v using the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
in this equation; x=v; a=1; b=20; c= -3000
v+=+%28-20+%2B-+sqrt%2820%5E2-4%2A1%2A-3000+%29%29%2F%282%2A1%29+
:
v+=+%28-20+%2B-+sqrt%28400-%28-12000%29+%29%29%2F2+
:
v+=+%28-20+%2B-+sqrt%2812400+%29%29%2F2+
Two solutions, we only want the positive solution
v+=+%28-20+%2B+111.355%29%2F2+
v = 91.355%2F2
v = 45.68 mph for a stopping distance of 150 ft
:
:
See if that flies in the original equation
d%28v%29=45.68%5E2%2F20%2B45.68
d%28v%29=2086.45%2F20%2B45.68
d%28v%29=104.32%2B45.68
d(v) = 150.00, confirms our solution