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Running A Yellow light

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Homework Statement
I was given this assignment today and I seriously don’t understand these questions
Homework Equations
When driving a car, a very real problem is presented by driving a little too fast. It can be impossible to get in the red. For these calculations two options will be considered. The first is to keep going at the same speed and the second is as fast as possible. A third option which involves accelerating through the intersection, and it is not safe to use. To simplify these calculations, reaction time will be ignored.

Mr. G is driving a car at 25m / s on a snowy road. With a traffic light that has a "time for yellow" of 3.0 s, how Far away from the light could the car be without breaking?
It has a maximum acceleration of -3.0 m/s^2 how many seconds will it be required to stop the car?
What is the average velocity of the car during this time that it is stopping in problem 2, and how does it go during this time?
Between what distances from the stop light will the car in a problem 2 have “no options” and run through the red light no matter what the driver does?Draw a sketch.
How many seconds would the "time for yellow" have to be for the car to be able to just get through the light at the constant velocity from the minimum stopping distance fund in problem 3?
I did the first three questions and found that the ignition distance without breaking would be 75 m, it would take 8.3 seconds to stop the car with a maximum acceleration of -3.0 m/s^2 and you would go 103,75 m if so.
 

rcgldr

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I get 104.17 (104 + 1/6) m for the stopping distance.
 

collinsmark

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Hello @Bellahhhh,

Welcome to PF! :welcome:

Homework Statement: I was given this assignment today and I seriously don’t understand these questions
Homework Equations: When driving a car, a very real problem is presented by driving a little too fast. It can be impossible to get in the red. For these calculations two options will be considered. The first is to keep going at the same speed and the second is as fast as possible. A third option which involves accelerating through the intersection, and it is not safe to use. To simplify these calculations, reaction time will be ignored.

Mr. G is driving a car at 25m / s on a snowy road. With a traffic light that has a "time for yellow" of 3.0 s, how Far away from the light could the car be without breaking?
It has a maximum acceleration of -3.0 m/s^2 how many seconds will it be required to stop the car?
What is the average velocity of the car during this time that it is stopping in problem 2, and how does it go during this time?
Between what distances from the stop light will the car in a problem 2 have “no options” and run through the red light no matter what the driver does?Draw a sketch.
How many seconds would the "time for yellow" have to be for the car to be able to just get through the light at the constant velocity from the minimum stopping distance fund in problem 3?
In the future, please use the provided template, show your relevant equations/formulae, and show your work.

By the way, I think you mean "braking," rather than "breaking."

I did the first three questions and found that the ignition distance without breaking would be 75 m, it would take 8.3 seconds to stop the car with a maximum acceleration of -3.0 m/s^2
Again, please show your work and the relevant equations. Also you might want to be careful about your precision (significant digits). Rounding errors that occur early on can propagate in later calculations.

and you would go 103,75 m if so.
I think one of your rounding errors might have propagated to this answer. Try that calculation again paying more attention to precision.
 

haruspex

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I concur that it seems you have rounding errors, but in particular that these have arisen by using the rounded numerical values of intermediate answers. Namely, you found the time to stop as 25/3, rounded to 8.3, and the average speed as 25/2=12.5, then multiplied these together.
A better process is to work entirely symbolically, only plugging in numbers to produce the answers requested. E.g. time to stop =v/a, average speed =v/2, distance covered =(v/a)(v/2)=v2/(2a)=625/6, rounded to 104.
 

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