Chapter 4.1, Problem 22PPS

(a)

To determine

To find: a linear equation in slope intercept form for distance of the car from G.

Answer to Problem 22PPS

  $y=15x+5$

Explanation of Solution

Given information:

G is driving a car at a constant speed. He starts the time when the car is 5 feet away. After 2 seconds the car was 35 feet away.

Calculation:

Let x be the seconds and y be the feet the car is away

In 0 seconds the car is 5 feet away.

  $\left(x_1,y_1\right)=\left(0,5\right)$

In 2 second the car is 35 feet away.

  $\left(x_2,y_2\right)=\left(2,35\right)$

The linear equation for distance in slope intercept form will be $y=mx+b$

Use the points $\left(0,5\right)$  and $\left(2,35\right)$

Slope can be calculated as shown:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{35-5}{2-0}\\ m=15\end{array}$

And y -intercept is 5

So the equation for distance of G from car becomes $y=15x+5$ .

(b)

To determine

To find: the distance travelled by the car after 10 seconds.

Answer to Problem 22PPS

The distance will be $155$ feet.

Explanation of Solution

Given information:

G is driving a car at a constant speed. He starts the time when the car is 5 feet away. After 2 seconds the car was 35 feet away.

Calculation:

The distance travelled by the car after 10 seconds can be calculated as

  $\begin{array}{c}y=15x+5\\ =15\left(10\right)+5\\ =150+5\\ =155\end{array}$

Therefore, the distance will be $155$ feet.