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.