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.
Your answer is d=_____
Estimate the stopping distance if the brakes are applied when the car is traveling at 62 miles per hour.
Your answer is_____.
3.If the point is on the unit circle in quadrant IV, then y=___.
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.
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