Chapter 4.2, Problem 22PPS

Solution

a.

A linear equation to find the distance of the car from Greg.

The linear equation to find the distance of the remote control car from Greg is $d\left(t\right)=5+15t$ .

Given: A person is driving the remote control car at constant speed. And he starts the timer when the car is 5 feet away. But, after 2 seconds the car is 35 feet away.

Calculation:

Considering at time $\left(t=0\right)$ the car is at a distance 5 feet and after 2 seconds $\left(t=2\right)$ the car is at a distance 35 feet.

Hence, the distance travelled by the car from time $\left(t=0\right)$ to $\left(t=2\right)$ is 35 feet-5 feet, i.e. 30 feet.

So, speed of the car is evaluated as,

  $\begin{array}{c}{\text{Car}}_{\text{speed}}=\frac{30}{2}\\ =15\text{ feet/sec}\end{array}$

Also, initially the car is 5 feet away. Therefore, the distance $d\left(t\right)$ of the remote control car from the Greg after $\left(t\right)$ seconds is written as,

  $d\left(t\right)=5+15t\mathrm{......}\left(1\right)$

b.

The distance travelled by the car after 10 seconds.

The distance travelled by the car after 10 seconds is 155 feet.

Given: A person is driving the remote control car at constant speed. And he starts the timer when the car is 5 feet away. But, after 2 seconds the car is 35 feet away.

Calculation:From (1), the distance travelled by the car after $\left(t\right)$ seconds is given as

  $d\left(t\right)=5+15t\mathrm{......}\left(2\right)$

As per the problem, the distance travelled by the remote control car after 10 seconds can be evaluated by plugging $\left(t=10\right)$ in $\left(2\right)$ . Therefore from (2)

  $\begin{array}{l}d\left(10\right)=5+15\left(10\right)\mathrm{......}\left(\text{From }\left(2\right)\right)\\ d\left(10\right)=155\text{ feet}\end{array}$