How Does Speed Affect Car Stopping Distance?

[HaLAA]

Homework Statement

the minimum distance for a car traveling at a speed of 30m/s is 60 m, including the distance traveled during the driver's reaction time of 0.44s. what is the minimum stopping distance for the same car traveling at the speed of 41m/s?

v=30m/s, s=60m
drivers' reaction time = 0.44s

2. The attempt at a solution
because the car traveling at a speed of 30m/s and the distance is 60m, so I find the traveling time 60m/30m/s=2s, and the driver's reaction time is 0.44s, thus the entire stopping time is 2s-0.44s=1.56s.

 

[BvU]

60m / 30 m/s is indeed 2 s. After, say 1.5 seconds, is the speed still 30 m/s ? In other words: can you really use the equation you are using in the case described ? Or do you need some other equations ?

Oh, and: wecome to PF :-)

Jsut so you know: using the template is mandatory. And the template has 1, 2 and 3. Especially 2 is of interest in your situation.

 

[HaLAA]

60m / 30 m/s is indeed 2 s. After, say 1.5 seconds, is the speed still 30 m/s ? In other words: can you really use the equation you are using in the case described ? Or do you need some other equations ?

Oh, and: wecome to PF :)

Jsut so you know: using the template is mandatory. And the template has 1, 2 and 3. Especially 2 is of interest in your situation.

the entire travling timie is 2s, I am sure. Then at the end the travel the speed I think is 0 m/s. So I think I can do

0=a*1.56s+30 m/s to fine the a which is -19m/s/s. Because it is the same car, I think the acceleration is the same.
thus, the distance at 41m/s, I think it should be 0.44s*41m/s+ 1/2*(-19m/s/s*(1.56s)²)+41m/s*1.56s

 

[BvU]

the entire traveling time is 2s, I am sure

No, it is not. Fill in 30 m/s instead of 41 m/s in your last expression and you don't get 60 m.

So, no template used, and still I have given some assistance. I hope I don't get banned from PF for doing so.
Perhaps I haven't made clear that, although I conceded that 60 m / 30 m/s = 2 s, that is the wrong equation to use.
That is also the argument for wanting you to use the template: That equation you don't provide applies to uniform motion with constant velocity, no acceleration therefore. So it does NOT apply to the situation after 0.44 s.

All remaining steps you make are impeccable: you can indeed assume that the deceleration for different speeds is the same and your last expression is just fine. Now find the right a.

 

[HalfLostAndSearching]

Using the formula s=ut+1/2at^2, I can calculate the minimum stopping distance for the same car traveling at a speed of 41m/s by substituting the values for velocity (u=41m/s), time (t=1.56s) and acceleration (a=-30m/s^2, as the car is decelerating). This gives a minimum stopping distance of 64.62m. Therefore, the minimum stopping distance for the car traveling at a speed of 41m/s is 64.62m.