Question

Greg car travels 13 feet per second. After 2 seconds his car was 30.7 feet from the start line . When then time started marti's cad was 2 feet in front of the start line . Her car travels 16.5 feet per second . Write an equation that represents the relationship between the time and distance

Answer 1

Answer:

The linear model of Greg's car is given by:

$d=13t+4.7$

The linear model of Marti's car is given by:

$d=16.5t+2$

Step-by-step explanation:

Let:

$d\rightarrow$ represent distance traveled by the cars in feet.

$t\rightarrow$ represent time of travel in seconds.

Greg's Car:

Given:

Rate of change = 13 ft/s

After 2 seconds his car was 30.7 feet.

From the given data we have the following:

Slope of line $m$ = 13 ft/s

Point $(2,30.7)$

Using point slope equation to model the linear relationship.

$y-y_1=m(x-x_1)$

where $(x_1,y_1)$ is a point on line.

So, we have:

$d-30.7=13(t-2)$

$d-30.7=13t-26$    [ Using distribution.]

Adding 30.7 to both sides.

$d-30.7+30.7=13t-26+30.7$

$d=13t+4.7$

The linear model of Greg's car is given by:

$d=13t+4.7$

Marti's car:

Given:

Rate of change = 16.5 ft/s

Starting point = 2 ft

From the given data we have the following:

Slope of line $m$ = 16.5 ft/s

y-intercept = 2 ft

Using slope intercept equation to model the linear relationship.

$y=mx+b$

where $m$ is slope of line or rate of change and $b$ is the y-intercept or stating point.

So, we have:

$d=16.5t+2$

The linear model of Marti's car is given by:

$d=16.5t+2$