The stopping distance of a car traveling 25mph is 61.7ft, and for a car traveling 35mph it is 106ft. The stopping distance in feet can be described by the equation y=ax^2 + bx + c, where x is the speed in mph.
Find the values of a and b.
import matplotlib.pyplot as plt
plt.plot([1,2,3,4])
plt.ylabel('some numbers')
plt.show()
To find the values of a and b in the equation y = ax^2 + bx + c, we need to use the given information about the stopping distances at different speeds.
Given:
When x = 25 (speed in mph), y = 61.7 (stopping distance in feet)
When x = 35 (speed in mph), y = 106 (stopping distance in feet)
We can set up two equations using these given data points:
Equation 1: 61.7 = a(25^2) + b(25) + c
Equation 2: 106 = a(35^2) + b(35) + c
To solve for a and b, we can solve this system of equations.
Let's start by simplifying the equations:
Equation 1: 61.7 = 625a + 25b + c
Equation 2: 106 = 1225a + 35b + c
Since there is a constant term (c) in both equations, we can subtract Equation 1 from Equation 2 to eliminate it:
Equation 2 - Equation 1: 106 - 61.7 = (1225a + 35b + c) - (625a + 25b + c)
44.3 = 600a + 10b
Now we have one equation with only the variables a and b.
Next, we can substitute one of the given data points into this equation to get another equation.
Let's use the first data point: (25, 61.7)
44.3 = 600a + 10b
61.7 = 625a + 25b + c
Now we have a system of two equations with two variables. We can solve this system of equations to find the values of a and b.
To find the values of a and b in the equation y = ax^2 + bx + c, we need to make use of the given information and solve a system of equations.
We are given two points on the graph: (25, 61.7) and (35, 106). We can use these points to set up two equations:
Equation 1: 61.7 = a(25)^2 + b(25) + c
Equation 2: 106 = a(35)^2 + b(35) + c
Let's simplify these equations:
Equation 1: 61.7 = 625a + 25b + c (1)
Equation 2: 106 = 1225a + 35b + c (2)
Now, we can subtract Equation 1 from Equation 2 to eliminate the constant term (c):
Equation 2 - Equation 1: 106 - 61.7 = (1225a - 625a) + (35b - 25b) + (c - c)
44.3 = 600a + 10b (3)
So, we have the following system of equations:
Equation 1: 61.7 = 625a + 25b + c
Equation 3: 44.3 = 600a + 10b
To solve this system, we can eliminate one variable. Let's multiply Equation 3 by 2:
2(44.3) = 2(600a + 10b)
88.6 = 1200a + 20b (4)
Now, we can subtract Equation 1 from Equation 4 to eliminate (c) as well:
Equation 4 - Equation 1: 88.6 - 61.7 = (1200a - 625a) + (20b - 25b) + (c - c)
26.9 = 575a - 5b
From this equation, we can solve for b:
-5b = 26.9 - 575a
b = (26.9 - 575a) / (-5) (5)
Now, substitute the expression for (b) from Equation (5) into Equation (3):
44.3 = 600a + 10((26.9 - 575a) / (-5))
Simplify this equation:
44.3 = 600a - 10(26.9 - 575a) / 5
Multiply both sides by 5 to eliminate the fraction:
221.5 = 3000a - 10(26.9 - 575a)
Expand and rearrange:
221.5 = 3000a - 269 + 5750a
Combine like terms:
221.5 = 8750a - 269
Add 269 to both sides:
221.5 + 269 = 8750a
Combine like terms:
490.5 = 8750a
Divide by 8750:
a = 490.5 / 8750
a = 0.056
Now that we have the value of a, we can substitute it back into Equation (5) to find b:
b = (26.9 - 575a) / (-5)
b = (26.9 - 575(0.056)) / (-5)
b ≈ -4.168
Therefore, the values of a and b are approximately a = 0.056 and b ≈ -4.168.
Well, you have two instances. Plug in the values and you'll have two equations in a and b. Solve them.
We know that when x is 0, y=0. (The car is not moving, so it has zero stopping distance.) So, c=0
61.7 = a(625)+b(25)
106 = a(1225)+b(35)
625a + 25b = 61.7
1225a + 35b = 106
multiply top by 7, bottom by 5, and subtract:
4375a + 175b = 431.9
6125a + 175b = 530
1750a = 98.1
a = 0.056
b = 1.067