Solve Critical Thinking Math: Stopping Dist. 0.0074v²

[JD003]

This is a homework assignment, I think I can work it out if I can set things up correctly. Do I to set up a quadratic equation in the form:

0 = ax + bx + c then factor for the zeros/solutions

(d) distance would be 0. But how do I put, (v) mph, with 0.0074v² ? More correctly asked I suppose is how do I separate it? If I simply plug v in, using #1. as the example, I see 0.0074(80)² and I know that's not getting me anywhere.

Thanks for any time you may have to help On a certain type of street surface, the equation d = 0.0074v² can be used to approximate the stopping distance d, in feet, a car traveling at a
speed of v miles per hour will need when its brakes are applied.

1. What is the cars stopping distance if its speed is 80 miles per hour?

The stopping distance would be ___ ft. (Round to nearest whole number.)

2. What is the cars stopping distance if its speed is 65 miles per hour?

The stopping distance would be ___ ft. (Round to nearest whole number.)

3. A car was in an accident. When the traffic officer got to the scene, he noticed the length of the skid mark left by the car. The officer measured the stopping
distance and concluded that driver need 41 ft. to stop after applying the brakes. At what speed was the car going?

The speed of the car was ___ mph.

4. Did the officer issue the car owner a ticket for speeding if the speed limit was 65 mph.

O A. No

O B. Yes

 

[MarkFL]

Is this problem part of a graded assignment, that is, does it contribute to your final grade?

The reason I ask is that our policy is not to provide help with graded work, as your professor most likely expects it to be your own work and we take academic honesty very seriously.

If this is a graded assignment, we can still help you with a similar problem from your textbook, so you can then gain an idea how to do the graded problem. :D

 

[JD003]

Ahh, yes it is graded. I'm not trying to circumvent the rules. I apologize that it does. I tried to phrase my question much like I would with one of the tutors at the library which is allowed at school at least. In that "I can figure it out of when I get the equation set up" I'm just not seeing how to do that.

We're not using a textbook. It's a hybrid class. 4 hrs a week classroom, the rest is on line in "Mylabs." When they do set up a "similar" example they just change the variables. In this case it would be the speed.

If I'm to far out of line, please feel free to delete the thread. And Again I apologize for bending or perhaps breaking the rules.Edit...Making progress. Turns out it's as easy as it looks in one respect.
In Questions 1 and 2 I'm just squaring (v), the speed and multiplying it by 0.0074. I'm getting correct answers

#1. 47 mph
#2. 31 mph

But I'm not impressed by my method. I still haven't drawn up the equation, I'm just calculating. Correct answers are enough for the grade but I'd like to set it up to solve for zero, which will serve me better down the road I'm sure.

I think the answer to #3 is 80 but again I don't see my method as proper.

 

[MarkFL]

It sounds to me like you are allowed to get outside help with our homework, if you are allowed to go to your school's tutors.

For the first two parts of the question, you are in fact only in need of plugging in for $v$ to find the value of $d$, since the equation you are given is solved for $d$.

For the third part, you are given $d$ and asked to find $v$, taking the positive root since $v$ is speed, which is non-negative. Can you solve the given equation for $v$?

 

[Deveno]

One thing to note is that the two equations:

$d = kv^2$ and:

$kv^2 - d = 0$

are equivalent. You *CAN* use the quadratic formula on the 2nd equation, but it seems to me simpler to just find that:

$v = \pm \sqrt{\dfrac{d}{k}}$

(note that speed is usually taken to be the absolute value of velocity, which avoids the pesky question of signs).