Question

1. cameron stops to get gas soon after beginning a road trip. he checks his distance from home 2 hours after filling his gas tank and checks again 5 hours after filling his gas tank. the first time he checked, he was 170 miles from home. the second time, he was 385 miles from home. what equation models cameron’s distance from home as a function of time since getting gas?

Explanation:

Step1: Identify the two points

We have two points in the form (time, distance): (2, 170) and (5, 385).

Step2: Calculate the slope (rate of change)

The slope \( m \) is calculated using the formula \( m=\frac{y_2 - y_1}{x_2 - x_1} \).
Substituting the values: \( m=\frac{385 - 170}{5 - 2}=\frac{215}{3}\approx71.67 \)? Wait, no, 385 - 170 is 215? Wait, 385 - 170 = 215? Wait, 170 + 215 = 385? Yes. And 5 - 2 = 3. Wait, but maybe I made a mistake. Wait, 385 - 170 is 215? Wait, 170 + 215 = 385. Yes. So slope \( m=\frac{215}{3}\approx71.67 \)? Wait, no, wait 385 - 170 is 215? Wait, 170 + 215 = 385. Yes. And 5 - 2 = 3. So \( m = \frac{385 - 170}{5 - 2}=\frac{215}{3}\approx71.67 \)? Wait, no, wait 385 - 170 is 215? Wait, 170 + 215 = 385. Yes. But maybe I miscalculated. Wait, 385 - 170: 385 - 100 = 285, 285 - 70 = 215. Yes. So slope is \( \frac{215}{3}\approx71.67 \)? Wait, no, wait 5 - 2 is 3. So \( m=\frac{215}{3}\approx71.67 \). Then use point - slope form. Let's use the point (2, 170). The point - slope form is \( y - y_1=m(x - x_1) \). So \( y - 170=\frac{215}{3}(x - 2) \). Let's convert it to slope - intercept form. \( y=\frac{215}{3}x-\frac{430}{3}+170 \). \( 170=\frac{510}{3} \), so \( y=\frac{215}{3}x-\frac{430}{3}+\frac{510}{3}=\frac{215}{3}x+\frac{80}{3}\approx71.67x + 26.67 \). Wait, but maybe I made a mistake in the slope. Wait, 385 - 170 is 215, 5 - 2 is 3. 215 divided by 3 is approximately 71.67. Alternatively, maybe the time is since getting gas. Wait, the problem says "time since getting gas". So at time \( t = 0 \) (when he gets gas), what's his distance? Wait, when \( t = 2 \), distance is 170; when \( t = 5 \), distance is 385. So the slope is \( \frac{385 - 170}{5 - 2}=\frac{215}{3}\approx71.67 \) miles per hour. Then the equation is \( d(t)=mt + b \). Using point (2, 170): \( 170=\frac{215}{3}(2)+b \). \( \frac{430}{3}+b = 170 \). \( b = 170-\frac{430}{3}=\frac{510 - 430}{3}=\frac{80}{3}\approx26.67 \). So the equation is \( d(t)=\frac{215}{3}t+\frac{80}{3} \), or we can write it as \( d(t)=71.67t + 26.67 \) (approximate) or in fraction form. Alternatively, maybe the slope is calculated as (385 - 170)/(5 - 2)=215/3≈71.67. So the linear equation is \( d(t)=71.67t + 26.67 \) or in exact form \( d(t)=\frac{215}{3}t+\frac{80}{3} \).

Wait, maybe I made a mistake. Let's re - calculate the slope. 385 - 170 = 215. 5 - 2 = 3. 215 divided by 3 is 71 and 2/3. So the slope \( m = \frac{215}{3} \). Then using the point (2, 170):

\( y - 170=\frac{215}{3}(x - 2) \)

\( y=\frac{215}{3}x-\frac{430}{3}+170 \)

\( 170=\frac{510}{3} \), so \( y=\frac{215}{3}x-\frac{430}{3}+\frac{510}{3}=\frac{215}{3}x+\frac{80}{3} \)

So the equation that models Cameron's distance from home as a function of time since getting gas is \( d(t)=\frac{215}{3}t+\frac{80}{3} \) (or approximately \( d(t)=71.67t + 26.67 \))

Answer:

The equation is \( d(t)=\frac{215}{3}t+\frac{80}{3} \) (or in slope - intercept form approximately \( d(t)\approx71.67t + 26.67 \))