1. ## urgent help with acceleration please

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

2. Originally Posted by asnxbbyx113
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.
From an initial speed v the time to come to a halt under constant decelleration a, is t=v/a., and the distance travelled is s=(v/2)t.

So plugging in the values given:

80=(v/2)(v/7) = v^2/14

so:

v = sqrt[(14)(80)]

RonL

3. Originally Posted by asnxbbyx113
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
Stoping distance is:

s = v t1 - v^2/(2a)

where t1 is the reaction time, and all the variables are in consistent units.

So convert the initial speed from km/hr to m/s and then plug in the other values.

RonL

4. 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.