40 km/h * 1000/3600 = 11.1 m/s initial Vi
goes for .25 sec *11.1 = 2.78 m before braking
average speed = 11.1/2 = 5.55 m/s
v = Vi + a t
a = -0.8
0 = 11.1 -0.8 t
t = 13.9 s (I have a hunch your a should be -8.0 m/s^2 not -0.8)
then d = Vi t + (1/2) a t^2
d = 11.1(13.9) - 0.4 (13.9)^2
d = 154 - 77.3 = 76.7 meters
76.7 + initial 2.78 = 79.5 meters
The speed limit in a school zone is 40km/h. A driver traveling at this speed sees a child run into the road 13m ahead of his car. He applies the breaks, and the car decelerates at uniform rate of 0.8m/s². If the driver's reaction times is 0.25s, will the car stop before hitting the child?
2 answers
Given:
Vo = 40km/h = 11.1 m/s.
d1 = 13m. = Stopping distance.
a = -8.0 m/s^2?.
t = 0.25s. = Reaction time.
d2 = V*t =11.1 * 0.25 = 2.78 m.
d = d1-d2 = 13 - 2.78 = 10.2 m = Required stopping distance.
V^2 = Vo^2 ^ 2a*d = 0.
11.1^2 - 5.56*d = 0,
d = 22.2 m = Stopping distance.
Vo = 40km/h = 11.1 m/s.
d1 = 13m. = Stopping distance.
a = -8.0 m/s^2?.
t = 0.25s. = Reaction time.
d2 = V*t =11.1 * 0.25 = 2.78 m.
d = d1-d2 = 13 - 2.78 = 10.2 m = Required stopping distance.
V^2 = Vo^2 ^ 2a*d = 0.
11.1^2 - 5.56*d = 0,
d = 22.2 m = Stopping distance.