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
The 3rd line from the bottom devided by the second line from the top added to the 5th line from the bottom when read backwards is Tuf's "official" final answer!