A car traveling at 91.0 km/h approaches the turnoff for a resturant 30.0 meters ahead. If the driver slams on the breaks whith an acceleration of -6.40 m/s^2, what will be her stopping distance? I'm having trouble putting this into a formula and making it work.
2006-10-10 13:16:57 · 4 answers · asked by swimmertommy 1 in Science & Mathematics ➔ Physics
Vf^2 = Vo^2 +2aX
X = Vf^2 - Vo^2 / 2a
= 0^2 - (25.3 m/s)^2 / 2(-6.40)
= -640 / -12.8
X = 50m
2006-10-10 13:39:35 · answer #1 · answered by Meilleur_que_toi 4 · 0⤊ 0⤋
well, I'd convert the kph into metres per second then I would probably use (final velocity)^2 = (initial velocity)^2 +2*acceleration*distance ; setting final velocity = 0.
Then just to be on the safe side I would calculate the time it took to stop and calculate the distance travelled during that time to see if both answers agreed.
2006-10-10 20:40:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The formula does not involve t. Convert k/h to m/s:
91 k/h = 25.2777 m/s
x = (V^2)/(2a) = (25.277^2)/(2*6.4) = 49.919 m
2006-10-10 20:47:53 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
s=ut+att=91t+6.4(6.4)tt=30
Calculate t
2006-10-10 20:21:12 · answer #4 · answered by rav 4 · 0⤊ 0⤋