# Math Help - some math problems

1. ## some math problems

1.Suppose p varies inversely as the square root of q. If p=-7 when q=9, what is p if q is 6?
p=_____

2.The stopping distance d of an automobile is directly proportional to the square of its speed v. A car required 50 feet to stop when its speed was 80 miles per hour. Find a mathematical model that gives the stopping distance d in terms of its speed v.
Estimate the stopping distance if the brakes are applied when the car is traveling at 62 miles per hour.

3.If the point is on the unit circle in quadrant IV, then y=___.

1.Suppose p varies inversely as the square root of q. If p=-7 when q=9, what is p if q is 6?
p=_____
We know that
p = (const)/sqrt(q)

Using the first point we get:
-7 = (const)/sqrt(9)

Thus (const) = (-7)*sqrt(9) = (-7)*3 = -21

Thus
p = -21/sqrt(q)

When q = 6:
p = -21/sqrt(6)

-Dan

3. I'm going to do them in the order I think they are the easiest!

3. If that point is on the unit circle centered at the origin then the distance to the origin from that point must be 1. So root(x² + y²) = 1

16/25 + y² = 1²
y² = 9/25
y = -3/5
(y must be negative as you want the answer in the 4th quadrant)

2.

d = kv² (where k is a constant)
50 = k80²
50 = 6400k
k = 1/320

So the model is therefore d = v²/320

For the second part just put v = 62 into that equation.

1. p = k/root(q)

So -7 = k / root(9)
So k = -+ 21

Using this value of k : p = +-21 / root(6)

Hope that helps.

2.The stopping distance d of an automobile is directly proportional to the square of its speed v. A car required 50 feet to stop when its speed was 80 miles per hour. Find a mathematical model that gives the stopping distance d in terms of its speed v.
Estimate the stopping distance if the brakes are applied when the car is traveling at 62 miles per hour.
d = (const)v^2

So
50 ft = (const)*(80 mi/h)^2

So (const) = (50 ft)/(80 mi/h)^2 = 1/128 ft h^2/mi^2. (Note how strange this unit is. We need to remember that the formula we will be using converts an initial speed in mi/h to a stopping distance in ft.)

Thus
d = v^2/128.

So for v = 62 mi/h, d = (62)^2/128 ft = 30.0313 ft.

-Dan

5. Oops! I never even noticed the feet/miles issue. Darn.

*slaps himself*

6. Wow, i was really off in number 2. For the equation i got like d=d*v/v. lol thanks. The last one i need help in. I can't figure out that bow tie triangle thing.

Wow, i was really off in number 2. For the equation i got like d=d*v/v. lol thanks. The last one i need help in. I can't figure out that bow tie triangle thing.
What question is that? I don't see it.

-Dan

8. Oh, I'm sorry. Number 3. this one:

3.If the point is on the unit circle in quadrant IV, then y=___.

9. I answered that one at the top of my post. I messed up the second question, but I'm pretty sure I got question 3 right.

10. No it's apparently wrong. I got that same answer but without the negative. So it's not 3/5 or -3/5

11. Originally Posted by TheBrain
I answered that one at the top of my post. I messed up the second question, but I'm pretty sure I got question 3 right.
You got it right. That's why I didn't post an answer for 3 myself.

-Dan

12. ok whatever. Thanks guys, your the best.

No it's apparently wrong. I got that same answer but without the negative. So it's not 3/5 or -3/5
What makes you think -3/5 is incorrect?

-Dan

14. I put it in as an answer for my homework website and it says it is wrong.