Problema Solution
A car traveling at a speed of vo = 52 m/s stops smoothly (that is, its deceleration is constant) over a distance of d = 146 m.
1What is its acceleration during the time it is stopping? (Be careful about the sign!)
a =
2How long (what amount of time) does it take for the car to come to a stop?
tstop =
3) After the car has gone 1/3 of the stopping distance, what is its speed?
v1/3 =
Answer provided by our tutors
v0 = 52 m/s
d = 146 m
V^2 = Vo^2+2ad
1. V=0
0=Vo^2+2ad
Solve for a:
a= -(Vo^2)/(2d)=-(52^2)/(2*146)=- 9.26 m/s^2
2.
V=Vo+at
0=Vo+at
t=-Vo/a=-52/-9.26= 5.62 s
3.
x=Vo*t+0.5at^2
x=52*5.62 + 0.5(-9.26)5.62^2=146 m (146/3= 48.67 m)
V^2=Vo^2+2ax
V=(Vo^2+2ax)^0.5=(52^2+2(-9.26)(48.67))^0.5= 42.46 m/s