# Minimum stopping distance

The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the drivers reaction time of 0.560 s.
what is the min. stopping distance for the same car traveling at a speed of 38.0 m/s?

tried to use this equation v^2 - u^2 = 2 a s
but then didnt know what to do.

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Doc Al
Mentor
Use that equation to figure out the car's acceleration. What distance did you use? (How far has the car traveled before the brakes are applied?)

43.2m is the real distance he first travels with the velocity of 30 m/s. because i subtracted 16.8 m from the 60 m.
so my a= 10.416

Doc Al
Mentor
Good! Now apply that--in reverse--to the second case, where the speed is 38 m/s, to find the new total stopping distance. (You'll have to calculate a new "reaction time distance".) The car's acceleration remains the same.

i dont know how to find a new reaction time distance??
i used this formula :
0^2 - 38^2 = 2(10.416) x

Doc Al
Mentor
i dont know how to find a new reaction time distance??
Do it the same way as before. How far does the car move before the brakes are applied?

38 x .56?
=21.28 m
but what do i do with that?

Doc Al
Mentor
You add it to the other distance. Understand what happens when the person wants to stop: (1) there's a time delay due to reaction time, so the car moves some distance before the brakes are even applied; (2) once the brakes are applied, the car accelerates to a stop.

The total stopping distance is the sum of both of these distances.

you add 21.28 to 60? to get 81.28m?

Doc Al
Mentor
you add 21.28 to 60? to get 81.28m?
No. 60m was the total stopping distance for the 30m/s case--nothing to do with the 38 m/s case!

Add 21.28 m to the distance you calculated for the de-accelerating phase of the motion.

to the 69.3?
69.3+21.28 = 90.58?
sorry if i leave really quick i have to run to class in about 2 mins!

Doc Al
Mentor
Yes.

THANK YOU!!! so so much