What is the minimum stopping distance for a car traveling at a speed of 40 m/s?

AI Thread Summary
The discussion revolves around calculating the minimum stopping distance for a car traveling at 40 m/s, given that it takes 60 m to stop from 30 m/s, including a reaction time of 0.5 seconds. The user attempts to find the stopping distance by calculating acceleration using two methods, leading to different results. The first method, using the kinematic equation, gives an acceleration of -10 m/s², resulting in a total stopping distance of 100 m. The second method incorrectly assumes a total time of 2 seconds, leading to confusion and an incorrect acceleration of -20 m/s². The user acknowledges the mistake in calculating the stopping time and seeks clarification on the correct approach.
habibclan
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Kinematics-- Stopping distance

Homework Statement



The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver's reaction time of 0.50 s. What is the minimum stopping distance for the same car traveling at a speed of 40 m/s?

Homework Equations



vf^2 = vi^2 + 2ad

a = (delta v)/ (delta t)


The Attempt at a Solution



Since the reaction time is 0.5 s, the distance traveled is 40 x 0.5= 20 m. Now, to find the time it takes to decrease the velocity to 0 m/s, one needs to find the acceleration whcih can be calculated in two ways and can someone please explain to me why the first way is incorrect!

Calculate Acceleration:
1) use equation

vf^2 = vi^2 + 2ad

I use the data for the initial velocity (first scenario) 30 m/s and the stopping distance 60 m, plug it into the eequation and i get -10 m/s^2. When I use this acceleration, vf (0 m/s) and vi (40 m/s) I can calculate d=80 m, which i add to the 20 m which it takes to react and the answer I get is 100 m.

1) However, if i use the equation, in which delta t= 2s - 0.5 s = 1.5 s. I subtracted the reaction time, the acceleration I get is a= (delta v)/(delta t) = (-40)/ (1.5) = -20 m/s^2. This acceleration is different from the acceleration calculated using the first equation. Can someone please tell me why this method is incorrect?

Thanks a lot!
 
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habibclan said:
However, if i use the equation, in which delta t= 2s - 0.5 s = 1.5 s. I subtracted the reaction time, the acceleration I get is a= (delta v)/(delta t) = (-40)/ (1.5) = -20 m/s^2. This acceleration is different from the acceleration calculated using the first equation. Can someone please tell me why this method is incorrect?

Thanks a lot!
you've gone wrong here with your statement that delta t = 2s.
Where did this come from? If you are attempting to calculate the stopping time during acceleration in the first part of the problem, which itself is not correct, you should also note that the stop time during acceleration in the second part is not the same.
 
Oh right. That delta 2 was a stupid mistake as I divide the initial velocity by the distance to get 2 s :S which was just out of habit. Thanks for the correction! I get it now!
 
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