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

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`

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`