Stopping distance w human reaction time

[afa]

Homework Statement

determine the stopping distance for a car with an intitial speed of 26.9 m/s and a human reaction time of 0.9s for an acceleration of -4m/s^2

Homework Equations

x=vt t=v/a x=x+vt+.5at^2

The Attempt at a Solution

I used the second equation to find total time by adding it to .9s then plugged that into equation 3 and added that to x of the first equation?? what am i doing wrong??

 

[stevenbhester]

Alright, you are going 26.9 m/s. When you stop, you take .9s before the deceleration takes place.

So, it's .9 seconds plus however much time the deceleration takes.

As such,
Alright, you are going 26.9 m/s. When you stop, you take .9s before the deceleration takes place.

So, it's .9 seconds plus however much time the deceleration takes.

As such,
(Original Velocity)/(Deceleration rate)=Total Deceleration Time
(26.9m/s)/(4m/s^2)=Total Deceleration Time
6.725 seconds= Total Deceleration Time

Stopping Distance=(Average Velocity)(Total Deceleration Time)
(.5)(26.9 m/s)(6.725 seconds)=90.45125 meters

Now, you know how much distance it takes to stop.
You have to add how much distance you covered before stopping.
(reaction time)(velocity during reaction time)=Distance traversed during reaction time
(.9 seconds)(26.9m/s)=24.21 meters

Add the two to get your answer.
24.21 meters + 90.45125 meters= 114.66125 meters

And that's your answer.

 

[afa]

thank you so much, you seem to be the most helpful, do you think you could help me out on some more?

 

[stevenbhester]

Certainly. How else would I postpone doing my chemistry work? And I haven't actually taken AP physics, so I suggest checking my answers that I give you. I just like math and am good at figuring stuff out.

 

[rwisz]

I would suggest that you first define your variables and units clearly in order to avoid confusion. In this case, the initial speed should be given in meters per second (m/s), the acceleration in meters per second squared (m/s^2), and the human reaction time in seconds (s). Additionally, you should specify what the final stopping distance is being measured in (meters, kilometers, etc.).

Once your variables and units are defined, you can use the equations of motion to solve for the stopping distance. The first equation, x=vt, can be used to find the distance traveled during the reaction time of 0.9 seconds. The second equation, t=v/a, can be used to find the total time it takes for the car to come to a complete stop. Finally, the third equation, x=x+vt+.5at^2, can be used to find the total stopping distance.

Therefore, your solution should look something like this:

Given:
Initial speed (v) = 26.9 m/s
Acceleration (a) = -4 m/s^2
Human reaction time (t) = 0.9 s

Using x=vt, the distance traveled during the reaction time is:
x = (26.9 m/s)(0.9 s) = 24.21 m

Using t=v/a, the total time it takes for the car to stop is:
t = (26.9 m/s)/(-4 m/s^2) = -6.725 s

Note: The negative sign indicates that the car is decelerating.

Using x=x+vt+.5at^2, the total stopping distance is:
x = 24.21 m + (26.9 m/s)(-6.725 s) + 0.5(-4 m/s^2)(-6.725 s)^2
x = 24.21 m - 180.53 m + 90.265 m = -66.055 m

Therefore, the stopping distance for a car with an initial speed of 26.9 m/s and a human reaction time of 0.9 s for an acceleration of -4m/s^2 is approximately 66.055 meters.

It is important to note that this solution assumes ideal conditions and does not take into account factors such as road conditions, tire grip, and human error, which can affect the actual stopping distance