Question 92889: I'm really glad you're here. Math is just not my thing, obviously, but I'm glad its yours. Here it is:
1. The approximate distance it takes to stop a car, based on the speed the car is traveling, is given in the following chart:
Miles per Hour Stopping Distance (ft)
25 62
35 106
45 161
50 195
55 228
65 306
a. Find the quadratic function based on the stopping distance for 55 mph and 65 mph and the fact that at 0 mph the stopping distance is 0 feet.
b. Use that equation to predict the stopping distance for 55 mph and 65 mph and compare them with the distances given in the chart.
c. If the equation continues to be valid for higher speeds, how many feet would it take to stop a drag racer that reaches a speed of 230 mph?
Thanks so much for your help! Let me know if you or someone else knows trig or calculus...that not my cup of tea either.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. The approximate distance it takes to stop a car, based on the speed the car is traveling, is given in the following chart:
Miles per Hour Stopping Distance (ft)
25 62
35 106
45 161
50 195
55 228
65 306
a. Find the quadratic function based on the stopping distance for 55 mph and 65 mph and the fact that at 0 mph the stopping distance is 0 feet.
You have two points: (55,228) and (65,306)
You are told the intercept is 0
EQUATION Form: y = ax^2+bx+c
You have three points so you have three equations:
0 = 0^2a + 0b + c ;;;So c=0
228 =55^2a + 55b +0
306 =65^2a + 65b + 0
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Now you have two equations with two unknowns so solve for a and for b:
Comment: I'm going to use a calculator matrix method to speed this up:
a = 0.0562237762 and b = 1.053146853
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So the EQUATION: f(x) = 0.056237762x^2+ 1.053146853x
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b. Use that equation to predict the stopping distance for 55 mph and 65 mph and compare them with the distances given in the chart
st.dis.= f(55) = 228.04 ft
st.dis.= f(65) = 306.06 ft.
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c. If the equation continues to be valid for higher speeds, how many feet would it take to stop a drag racer that reaches a speed of 230 mph?
st.dis =f(230) = 3217.2 ft
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Thanks so much for your help! Let me know if you or someone else knows trig or calculus...that not my cup of tea either.
I suggest you post your problems. I'm sure amoung the 600 responders someone
knows trig and/or calculus.
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Cheers,
Stan H.
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