in coming to a stop, a car leaves skid marks 80m long on the highway. assuming a deceleration of 7.0m/s^2, estimate the speed of the car just before breaking.
determine the stopping distances for an automobile with an initial speed of 90km/h and human reaction time of 1.0s: a) for an accelerrtion a=-4.0m/s^2; b) for a=-8.0m/s^2
in coming to a stop, a car leaves skid marks 80m long on the highway. assuming a deceleration of 7.0m/s^2, estimate the speed of the car just before breaking.
Given:
distance, d = 80 m
deceleration, a = -7 m/sec/sec
Final velocity, Vf = 0
Initial velocity, Vo = unknown
Find: Vo
d = (average velocity)*(time)
80 = [(Vo +Vf) /2]*t
80 = [Vo +0](t/2)
80 = (Vo)t/2 ---------------(1)
Vf = Vo +at
0 = Vo -7t
t = (Vo)/7 ------------(2)
Substitute that in (1),
80 = (Vo)[(Vo)/7]/2
80 = [(Vo)^2 ]/14
(Vo)^2 = 80(14)
Vo = sqrt(80*14) = 33.47 m/sec ----the speed of the car just before breaking.
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determine the stopping distances for an automobile with an initial speed of 90km/h and human reaction time of 1.0s: a) for an accelerrtion a=-4.0m/s^2; b) for a=-8.0m/s^2
Human reaction time?
Oh, it means what is the stopping distance from the moment the driver sees the danger or needs to stop immediately.
So two distances are involved:
---1) the distance the car travels while it is still running at 90 kph. This is for 1 second.
---2) the distance the car travels the moment the brakes are hit. This is fot t seconds.
Given:
Vf = 0
Vo = 90kph = (90km/hr)(1000m/1km)(1hr/3600sec) = 25 m/sec
a = -4 m/sec/sec
d = ?
d = [(Vf +Vo)/2]*t
d = [0 +25](t/2)
d = 12.5t -------------------(i)
Vf = Vo +at
0 = 25 -4t
t = 25/4 = 6.25 sec
So,
d = 12.5(6.25) = 78.125 m ------------for the braking.
For the reaction time of 1 sec,
D = (speed)(time)
D = (25)(1) = 25 m
Therefore, the stopping distance is 78.125 +25 = 103.125 meters. ---answer.