On a dry road, a car with good tires may be able to brake with a constant deceleration of 3.65 m/s2. (a) How long does such a car, initially traveling at 26.3 m/s, take to stop? (b) How far does it travel in this time?
2007-01-28 05:47:48 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
I'd say 7.21 seconds and about 108.73 meters.
2007-01-28 05:57:29 · answer #1 · answered by Christopher C 3 · 0⤊ 0⤋
5
2007-01-28 13:51:05 · answer #2 · answered by odandme 6 · 0⤊ 0⤋
v2 = v1 + at
When the car is stopped v2 = 0, we know v1=26.3 and a = -3.65
0 = 26.3 -3.65(t)
t= 7.21 s
s2 = at^2 + v1t + s1
Solve for s2, say s1 is 0 (the starting point), v1, initial velocity = 26.3, t = 7.21, a = -3.65
s2 = .119 m
That second part was kinda weird.
2007-01-28 13:59:57 · answer #3 · answered by spidermilk666 6 · 0⤊ 0⤋
a) Vf=at + Vo
0=(-3.65)t + 26.3
t = 7.2 sec
b) Vf^2 - Vo^2 = 2ad
0^2 - 26.3^2 = 2 (-3.65) D
D = 94.8 m
2007-01-28 13:59:45 · answer #4 · answered by Dennis H 4 · 0⤊ 0⤋