A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 36 ft/s^2. What is the distance covered before the car comes to a stop?
2006-11-27 06:36:23 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
There has been one careless conversion of 50 mph to not quite the correct number of ft/sec. Remember that 60 mph = 88 ft/sec exactly.
S = V^2 / 2A, and V = 88 * 50 / 60, so S = 74.691. . . feet.
2006-11-27 08:42:10 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Just convert 50 miles/h into feet/s so 50 x 5280/3600s = 73.3 ft/s so then use formula Vf^2=Vi^2 + 2ad , then go 0ft/s=73.3ft/s^2 +2(36)d
d = 5372.89/2(36)
d=74.62ft
2006-11-27 14:56:19 · answer #2 · answered by creeping_death687 1 · 1⤊ 1⤋
About 74.7ft.
2006-11-27 14:43:54 · answer #3 · answered by JJ 7 · 1⤊ 0⤋
46.03748 feet.
2006-11-27 14:39:25 · answer #4 · answered by Anonymous · 0⤊ 1⤋