(adsbygoogle = window.adsbygoogle || []).push({}); Kinematics-- Stopping distance

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

vf^2 = vi^2 + 2ad

a = (delta v)/ (delta t)

3. The attempt at a solution

Since the reaction time is 0.5 s, the distance travelled 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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# Homework Help: Kinematics- Stopping distance

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