The police department must determine a safe speed limit on a bridge so that the flow rate of cars is at a maximum per unit time. The greater the speed limit, the farther apart the cars
must be expected to be in order to allow for a safe stopping distance. The total distance needed for a car to stop, if a car in front of it stops suddenly, depends on the speed of the car through two factors: the time needed to react and the braking distance. Experimental data on the braking distance d (in feet), on the bridge surface, for various speeds s (in miles per hour) is given in the
following table. The table also provides an estimate for reaction distance r (in feet); this is the
distance the car will travel before the driver reacts.
s (in mph) 5 10 20 30 40 50 60
d (in feet) 4 11 33 62 100 149 203
r (in feet) 5 10 20 30 40 50 60
The police department has also identified the lengths of the 5 most common types of vehicles that are expected to use the bridge:
Model Length (in inches)
Fiat 500 142.0
Ford Fiesta 153.1
Dodge Caliber 173.8
Honda Civic 176.5
Dodge Grand Caravan 202.5
The bridge will also be used occasionally by tractor-trailers with an average length of 75 feet and stopping distances that are about 40% greater than the stopping distances for an automobile.
1. Find a function of the form d(s) = as2 +bs+c that models the breaking distance in terms of speed s. What should d(0) equal? What is a reasonable estimate for d�Œ(0)? Select the constants a, b and c that best fits the data and produces the value of d(0) and d�Œ(0) that
you identified. Also find a function r(s) that models the reaction distance in terms of the
speed.
Please Help!!!!!
That's a lot of spurious data for answering one simple question. Something tells me there are followup questions coming. For now, all we are interested in is speed and BRAKing distance. (If the braking distance is too large, then we will get into the breaking distance!)
s (in mph) 5 10 20 30 40 50 60
d (in feet) 4 11 33 62 100 149 203
Letting d = 1.5(s/5)^2 gets us
s/5 (in mph) 1 2 4 6 8 10 12
d (in feet) 1.5 6 24 54 96 150 216
It underestimates for small s, and starts to overestimate at about s=50
You can play around with other terms, or use quadratic regression or divided differences.
Others may offer further insights.
You might notice that the slope increases by about 10 in each interval of 10. That is, the slope is just about the value of s.
d' = s
and then integrate that twice to get a quadratic and evaluate a,b,c.
To find a quadratic function of the form d(s) = as^2 + bs + c that models the braking distance, we can use the given data points in the table.
Since the function is quadratic, we need three points to determine the constants a, b, and c. Let's choose the points (20, 33), (40, 100), and (60, 203) from the table.
Using these points, we can set up a system of equations:
1. For the point (20, 33):
33 = a(20)^2 + b(20) + c
2. For the point (40, 100):
100 = a(40)^2 + b(40) + c
3. For the point (60, 203):
203 = a(60)^2 + b(60) + c
We also know that the reaction distance function r(s) is linear, given by r(s) = ks, where k is a constant.
Now let's solve the system of equations and find the constants a, b, c, and k.
1. For the point (20, 33):
33 = 400a + 20b + c. Equation A
2. For the point (40, 100):
100 = 1600a + 40b + c. Equation B
3. For the point (60, 203):
203 = 3600a + 60b + c. Equation C
To find a, b, and c, we can subtract Equation A from Equation B to eliminate c, and then subtract Equation B from Equation C:
Equation B - Equation A:
100 - 33 = 1600a + 40b + c - (400a + 20b + c)
67 = 1200a + 20b. Equation D
Equation C - Equation B:
203 - 100 = 3600a + 60b + c - (1600a + 40b + c)
103 = 2000a + 20b. Equation E
Now we have two equations (Equation D and Equation E) with two variables (a and b):
67 = 1200a + 20b. Equation D
103 = 2000a + 20b. Equation E
Subtract Equation D from Equation E to eliminate b:
103 - 67 = 2000a + 20b - (1200a + 20b)
36 = 800a
Divide both sides by 800:
a = 36/800
a = 0.045
Substitute the value of a back into Equation D or Equation E to find b:
67 = 1200(0.045) + 20b
67 = 54 + 20b
13 = 20b
b = 13/20
b = 0.65
Now, substitute the values of a and b into Equation A to find c:
33 = 400(0.045)^2 + 20(0.65) + c
33 = 0.81 + 13 + c
33 = 13.81 + c
c = 33 - 13.81
c = 19.19
So, the quadratic function that models the braking distance is:
d(s) = 0.045s^2 + 0.65s + 19.19
To find the value of d(0), we plug in s = 0 into the function:
d(0) = 0.045(0)^2 + 0.65(0) + 19.19
d(0) = 0 + 0 + 19.19
d(0) = 19.19
A reasonable estimate for d�Œ(0) would be the derivative of the function, which represents the rate of change of the braking distance with respect to speed. Taking the derivative of the function d(s), we get:
d�Ō(s) = 0.09s + 0.65
To find d�Ō(0), we plug in s = 0 into the derivative:
d�Ō(0) = 0.09(0) + 0.65
d�Ō(0) = 0 + 0.65
d�Ō(0) = 0.65
So, a reasonable estimate for d�Œ(0) is 0.65.
For the reaction distance function r(s), we were given that it is linear and can be represented as r(s) = ks, where k is a constant. We can use the given data points (5, 5), (10, 10), and (20, 20) from the table to solve for k.
Using the point (5, 5), we can set up the equation:
5 = 5k
k = 1
So, the function r(s) that models the reaction distance is:
r(s) = s
To find a function that models the braking distance in terms of speed, we need to use the given data to determine the values of the constants a, b, and c in the quadratic equation d(s) = as^2 + bs + c.
First, let's analyze the given data:
s (in mph): 5 10 20 30 40 50 60
d (in feet): 4 11 33 62 100 149 203
From the given data, we can form the following equations:
1. Using d(5) = 4:
4 = 25a + 5b + c (equation 1)
2. Using d(10) = 11:
11 = 100a + 10b + c (equation 2)
3. Using d(20) = 33:
33 = 400a + 20b + c (equation 3)
We can solve this system of equations to find the values of a, b, and c.
Step 1: Rearrange the equations to isolate the constants:
Equation 1: 25a + 5b + c = 4
Equation 2: 100a + 10b + c = 11
Equation 3: 400a + 20b + c = 33
Step 2: Use any method (substitution or elimination) to solve the system of equations. Here, we'll use the elimination method.
Subtract equation 1 from equation 2:
100a + 10b + c - (25a + 5b + c) = 11 - 4
75a + 5b = 7 (equation 4)
Subtract equation 1 from equation 3:
400a + 20b + c - (25a + 5b + c) = 33 - 4
375a + 15b = 29 (equation 5)
Step 3: Solve equations 4 and 5 simultaneously.
Multiply equation 4 by 5 and equation 5 by 1:
375a + 25b = 35 (equation 6)
375a + 15b = 29 (equation 5)
Subtract equation 5 from equation 6:
375a + 25b - (375a + 15b) = 35 - 29
10b = 6
b = 6/10
b = 0.6
Substitute b = 0.6 into equation 4:
75a + 5(0.6) = 7
75a + 3 = 7
75a = 7 - 3
75a = 4
a = 4/75
Now, substitute the values of a and b into equation 1 to find c:
25a + 5b + c = 4
25(4/75) + 5(0.6) + c = 4
4/3 + 3 + c = 4
c = 4 - (4/3 + 3)
c = 4 - (12/3 + 9/3)
c = 4 - 21/3
c = -5/3
Therefore, the equation that models the braking distance is:
d(s) = (4/75)s^2 + 0.6s - 5/3
To find a function r(s) that models the reaction distance in terms of speed, we can use the given data for r (in feet) and s (in mph) in a similar manner as above.
Using the given data:
s (in mph): 5 10 20 30 40 50 60
r (in feet): 5 10 20 30 40 50 60
We can observe that the reaction distance is directly proportional to the speed. So, we can write the equation as:
r(s) = ks
To find the constant k, we can use any data point, such as r(5) = 5:
5 = k(5)
k = 1
Therefore, the function that models the reaction distance is:
r(s) = s
As for the values of d(0) and r(0), we can evaluate the respective functions at s = 0:
d(0) = (4/75)(0)^2 + 0.6(0) - 5/3
= 0 - 0 + 0
= 0
r(0) = 0
So, the reasonable estimate for d(0) is 0, and the value of r(0) is 0 as well.