Accidental Reconstruction Post-Investigation (Physics)

[darealkb1]

Read the attached File

The radical equation below can be used as a model how distance of a skid mark is related to speed when the driver stopped, where v represents velocity/speed, d represents distance/length of skid mark, and k is the drag factor. Use this formula to calculate the speed of the federal vehicle when it stopped.

v = d over k ( d over k is a fraction)

*The accident reconstruction/investigation determined that the skid marks of the federal vehicle were 33 ft, 4 inches. The posted speed limit in the parking lot is 15mph. A drag sled test determined then drag factor to be 0.037

I need this problem worked out please. I'm frustrated! Thanks

 

[Dr.Peterson]

The radical equation below can be used as a model how distance of a skid mark is related to speed when the driver stopped, where v represents velocity/speed, d represents distance/length of skid mark, and k is the drag factor. Use this formula to calculate the speed of the federal vehicle when it stopped.

v = d over k ( d over k is a fraction)

*The accident reconstruction/investigation determined that the skid marks of the federal vehicle were 33 ft, 4 inches. The posted speed limit in the parking lot is 15mph. A drag sled test determined then drag factor to be 0.037

I need this problem worked out please. I'm frustrated! Thanks

Here is the entire content of the pdf, which could have been attached as an image, to save people having to open it:



The equation as you stated it is not a radical equation; it should be \(v=\sqrt{\frac{d}{k}}\), which you could type as v = sqrt(d/k).

What you haven't done is to show us where you are confused. All you have to do is plug in values for d and k; d has to be converted from feet and inches to whatever unit is appropriate, presumably miles. (The problem should state the units for the variables in the formula, or at least for the drag factor, from which you could determine the appropriate units). So if what units to use is your issue, then it is the author's fault. If it's how to do the conversion, we can help you best if we see how you tried to do it (so we can help with the method you know rather than one you don't).

 

[Deleted member 4993]

Read the attached File

The radical equation below can be used as a model how distance of a skid mark is related to speed when the driver stopped, where v represents velocity/speed, d represents distance/length of skid mark, and k is the drag factor. Use this formula to calculate the speed of the federal vehicle when it stopped.

v = d over k ( d over k is a fraction)

*The accident reconstruction/investigation determined that the skid marks of the federal vehicle were 33 ft, 4 inches. The posted speed limit in the parking lot is 15mph. A drag sled test determined then drag factor to be 0.037

I need this problem worked out please. I'm frustrated! Thanks

The problem-statement asks:

"..... calculate the speed of the federal vehicle when it stopped"

Speed of a vehicle when it is stopped = 0 . That is the definition of being "stopped".

 

[darealkb1]

Here is the entire content of the pdf, which could have been attached as an image, to save people having to open it:

View attachment 25081

The equation as you stated it is not a radical equation; it should be \(v=\sqrt{\frac{d}{k}}\), which you could type as v = sqrt(d/k).

What you haven't done is to show us where you are confused. All you have to do is plug in values for d and k; d has to be converted from feet and inches to whatever unit is appropriate, presumably miles. (The problem should state the units for the variables in the formula, or at least for the drag factor, from which you could determine the appropriate units). So if what units to use is your issue, then it is the author's fault. If it's how to do the conversion, we can help you best if we see how you tried to do it (so we can help with the method you know rather than one you don't).

Thank you for showing me how to attach a image. That was better. I totally agree with you about the equation part. I was confused about how to solve the equation. Please don't beat me up over my confusion. Now you have instructed me so I'm understanding it more. I can't reinvent the question because I have to answer the question rather than be negative how the equation was made. Thanks again!

 

[Dr.Peterson]

Thank you for showing me how to attach a image. That was better. I totally agree with you about the equation part. I was confused about how to solve the equation. Please don't beat me up over my confusion. Now you have instructed me so I'm understanding it more. I can't reinvent the question because I have to answer the question rather than be negative how the equation was made. Thanks again!

I'll make two more comments:

First, you should be aware that, technically, you are not "solving an equation", but just "evaluating an expression" (applying a formula).

Second, and much more important, my point was not so much to criticize you or the problem, as to point out that because it doesn't state what units are to be used, we can't be sure that your answer is correct if you do it as I suggested, though I think it is most likely right. If I were you, and if it matters much, I would ask my instructor whether it is true that the distance in the formula has to be in miles.

 

[darealkb1]

I will reply to you. This teacher is my daughter's teacher. I'm a parent. I am aware that the equation is faulty because I asked a forensic accident engineer and he said the equation was not radical but wrong. The teacher said don't worry yourself converting. Why? I don't know. I'm just trying to get the the problem solved. Boom! That's all. No aggression! No nonsense! Yes, she left lots of details to make anyone say, "This equation is not correct". I'm just trying to sift through the madness. Please don't come at me saying I don't see how or why your confused when its not a accurate equation to solve. I was only trying to ask for help which you did help me but your tone of your response would have anyone wanting to speak negative to you. I'm not going there. Thanks for your help.?