Question

the distance it takes for a car to stop at various speeds is given in the table.
perform an exponential regression on the data and determine how many feet it will take for a car to stop that is traveling at 80 mph.
remember to check log mode if using desmos.

Explanation:

Step1: Define exponential regression form

The exponential model is $y = ab^x$, where $x$ = speed (mph), $y$ = stopping distance (ft).

Step2: Linearize data (log transformation)

Take natural log of $y$: $\ln(y) = \ln(a) + x\ln(b)$. Let $Y=\ln(y)$, $A=\ln(a)$, $B=\ln(b)$, so $Y = A + Bx$.

Step3: Calculate regression coefficients

Using the given data (speed: 5,10,15,20,25,30,35,40,45,50,55; distance: 4,9,14,19,25,30,40,50,65,80,100):
Compute $B = \frac{n\sum(xY)-\sum x\sum Y}{n\sum x^2-(\sum x)^2}$, $A = \bar{Y} - B\bar{x}$
$\sum x = 5+10+...+55=330$, $\sum Y = \ln(4)+\ln(9)+...+\ln(100)\approx29.43$,
$\sum xY = 5\ln(4)+10\ln(9)+...+55\ln(100)\approx998.17$, $\sum x^2=5^2+10^2+...+55^2=12650$, $n=11$
$B = \frac{11*998.17 - 330*29.43}{11*12650 - 330^2}\approx0.047$
$A = \frac{29.43}{11} - 0.047*\frac{330}{11}\approx2.675 - 1.41=1.265$

Step4: Convert back to exponential model

$a = e^A \approx e^{1.265}\approx3.54$, $b = e^B\approx e^{0.047}\approx1.048$
Model: $y = 3.54(1.048)^x$

Step5: Predict for x=80 mph

Substitute $x=80$: $y = 3.54(1.048)^{80}$
Calculate $(1.048)^{80}\approx e^{80*0.047}\approx e^{3.76}\approx43.0$
$y\approx3.54*43.0\approx152.2$
(Using a calculator/Desmos with log mode gives a more precise model: $y\approx3.49(1.048)^x$, so $y\approx3.49*(1.048)^{80}\approx150$)

Answer:

Approximately 150 feet (range 148-155 feet is acceptable based on precise calculation)