Question
The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. If a car traveling 40 mph can stop in 100 ft, how fast can a car travel and still stop in 256 ft?
Answers
Just translate ...
d = k(r^2) , where k is a constant
plug in the given to find k
100/5280 = k(1600)
k = (100/1600)/(5280) = 1/84480
so your formula is
d = (1/84480)r^2
if d = 256 ...
256/5280 = (1/84480)r^2
84480(256)/5280 = r^2
256(16) = r^2
r =16(4)
= 64 mph
notice we could have skipped the 5280 ft to mile conversion, since in both cases the same units were used. e.g. mph <---> ft
.........................
we could have used a simple ratio:
d1/d1 = (r1)^2/(r2)^2
100 ft/256 ft = (40 mph)^2/ (r2)^2
25/64 = 1600/r2^2
take √ of both sides
5/8 = 40/r2
5r2 = 320
r2 = 64
d = k(r^2) , where k is a constant
plug in the given to find k
100/5280 = k(1600)
k = (100/1600)/(5280) = 1/84480
so your formula is
d = (1/84480)r^2
if d = 256 ...
256/5280 = (1/84480)r^2
84480(256)/5280 = r^2
256(16) = r^2
r =16(4)
= 64 mph
notice we could have skipped the 5280 ft to mile conversion, since in both cases the same units were used. e.g. mph <---> ft
.........................
we could have used a simple ratio:
d1/d1 = (r1)^2/(r2)^2
100 ft/256 ft = (40 mph)^2/ (r2)^2
25/64 = 1600/r2^2
take √ of both sides
5/8 = 40/r2
5r2 = 320
r2 = 64
Where does the 5280 come from?