Alcohol and Driving The concentration of alcohol in a person’s bloodstream is measurable.
Suppose that the relative risk R of having an accident while driving a car can be modeled by the
equation
R = e
kx
where x is the percent of concentration of alcohol in the bloodstream and k is a constant.
(a) Suppose that a concentration of alcohol in the bloodstream of 0.03 percent results in a relative risk
of an accident of 1.4. Find the constant k in the equation.
(b)Using the same value of k, what concentration of alcohol corresponds to a relative risk of 100?
(c) If the law asserts that anyone with a relative risk of having an accident of 5 or more should not
have driving privileges, at what concentration of alcohol in the bloodstream should a driver be
arrested and charged with a DUI?
(a) To find the constant k, we can use the information provided. We know that at a concentration of alcohol of 0.03 percent, the relative risk of an accident is 1.4. We can plug these values into the equation.
R = e^(kx)
1.4 = e^(0.03k)
To isolate k, we can take the natural logarithm (ln) of both sides:
ln(1.4) = ln(e^(0.03k))
Using the property of logarithms that ln(e^x) = x, we have:
ln(1.4) = 0.03k
Now we can solve for k by dividing both sides by 0.03:
k = ln(1.4) / 0.03
Using a calculator, we find that k is approximately 0.154.
(b) To find the concentration of alcohol that corresponds to a relative risk of 100, we can plug this value into the equation and solve for x.
R = e^(kx)
100 = e^(0.154x)
Taking the natural logarithm of both sides:
ln(100) = ln(e^(0.154x))
Simplifying using the property of logarithms:
ln(100) = 0.154x
Now we can solve for x by dividing both sides by 0.154:
x = ln(100) / 0.154
Using a calculator, we find that x is approximately 2.416 percent.
(c) If the law asserts that anyone with a relative risk of having an accident of 5 or more should not have driving privileges, we can set the equation equal to 5 and solve for x.
R = e^(kx)
5 = e^(0.154x)
Taking the natural logarithm of both sides:
ln(5) = ln(e^(0.154x))
Using the property of logarithms:
ln(5) = 0.154x
Now we can solve for x:
x = ln(5) / 0.154
Using a calculator, we find that x is approximately 0.923 percent. Therefore, at a concentration of alcohol in the bloodstream of approximately 0.923 percent or higher, a driver should be arrested and charged with a DUI.
To solve the problem, we need to find the constant "k" in the equation R = e^(kx) and use it to solve parts a, b, and c.
(a) To find the constant "k" when x = 0.03 and R = 1.4, we can substitute these values into the equation:
1.4 = e^(0.03k)
Now, we need to solve for "k". Taking the natural logarithm (ln) of both sides of the equation, we get:
ln(1.4) = ln(e^(0.03k))
Using the property of logarithms, ln(e^(0.03k)) simplifies to 0.03k:
ln(1.4) = 0.03k
Divide both sides of the equation by 0.03:
k = ln(1.4) / 0.03
Using a calculator or computer, calculate the value of k to obtain the answer.
(b) To find the concentration of alcohol (x) that corresponds to a relative risk (R) of 100, we can use the same value of k obtained in part (a). Substitute the values into the equation:
100 = e^(0.03k)
We now need to solve for "x". Taking the natural logarithm of both sides:
ln(100) = ln(e^(0.03k))
Using the property of logarithms, ln(e^(0.03k)) simplifies to 0.03k:
ln(100) = 0.03k
Divide both sides by 0.03:
k = ln(100) / 0.03
Now that we have the value of k, we can calculate the corresponding concentration of alcohol by substituting it back into the equation.
(c) To determine the concentration of alcohol in the bloodstream at which a driver should be arrested and charged with a DUI (having a relative risk of 5 or more), we need to find the corresponding value of "x" when R = 5.
Substitute R = 5 and the value of k obtained in part (a) into the equation:
5 = e^(0.03k)
By taking the natural logarithm of both sides:
ln(5) = ln(e^(0.03k))
Using the property of logarithms, ln(e^(0.03k)) simplifies to 0.03k:
ln(5) = 0.03k
Divide both sides by 0.03:
k = ln(5) / 0.03
Now that we have the value of k, we can calculate the concentration of alcohol corresponding to a relative risk of 5 by substituting it back into the equation.
Remember to use a calculator or computational software to perform any calculations involving natural logarithms and exponentials.