highway safety engineers can use the formula
d= 0.05s^2+1.1s to estimate the minimum stopping distance d in the feet for a vehicle traveling s miles per hour. If a car is able to stop after 125 ft. what is the fastest it could have been traveling when the driver first sppled the brakes?
2007-11-26 08:50:02 · 4 answers · asked by andy l 1 in Science & Mathematics ➔ Mathematics
d = 0.05s^2+1.1s <<--- ???? was this given to u ?
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heres the formula
but i am lost of what u gave me in your data !
dfinal = 0.5 * g * t² + V(t) + dinitial
Vfinal = g * t² +Vinitial
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Lenpol7 got it ---good man
i gave you a thumbs up
too much medication today
2007-11-26 09:04:30 · answer #1 · answered by JavaScript_Junkie 6 · 1⤊ 1⤋
This is just a simple quadratic equations:
125 = 0.05s² + 1.1s
or 0.05s² + 1.1s - 125 = 0
Using the quadratic equation, we can get
s = (-b ± â(b² - 4ac))/2a = (-1.1 ± â(1.21 - 4 * 0.05 * (-125)))/(2*0.05)
s = 40.2 mph.
(The other solution to this equation is s = -62.2 mph, but I think we can disallow negative speeds in this case.)
2007-11-26 17:10:38 · answer #2 · answered by dansinger61 6 · 0⤊ 0⤋
d = 0.05s^2 + 1 & re-arranging
0.05s^2 = d - 1
s^2 = (d - 1) / 0.05
s = sq rt[(d - 1) / 0.05]
s = sq rt[(125 - 1) / 0.05]
s = sq rt[ 124/ 0.05]
s = sq rt[ 2480]
s = 49.79959 mph
s = 49.8 mph
2007-11-26 17:02:15 · answer #3 · answered by lenpol7 7 · 1⤊ 1⤋
i might have the answer tomorrow
2007-11-26 16:58:13 · answer #4 · answered by Gabrielle S 2 · 0⤊ 1⤋