Question

1. sketch the graph, then answer the question below. a. kyler starts 2 feet away from the motion detector. b. he then walks away from the motion detector at a rate of 2 feet per second for 2 seconds. c. he remains still for 2 seconds. d. he then walks toward the motion detector at a rate of - 1/3 ft/sec for 6 seconds. how far is kyler from the motion detector when he finishes his walk? kyler is ____ ft away from the motion detector when he finishes.

Explanation:

Step1: Analyze part - a

Kyler starts 2 feet away from the motion detector. For the first part of his motion (0 - 2 seconds), he walks away at a rate of 2 feet per second. The distance $d$ as a function of time $t$ (for $0\leq t\leq2$) is $d(t)=2 + 2t$. At $t = 2$ seconds, $d(2)=2+2\times2=6$ feet.

Step2: Analyze part - b

From $t = 2$ to $t = 4$ seconds, he walks away at a rate of 2 feet per second. The distance function for $2\lt t\leq4$ is $d(t)=6 + 2(t - 2)=2t + 2$. At $t = 4$ seconds, $d(4)=2\times4+2 = 10$ feet.

Step3: Analyze part - c

From $t = 4$ to $t = 6$ seconds, he remains still. So the distance function for $4\lt t\leq6$ is $d(t)=10$ (a horizontal - line on the graph).

Step4: Analyze part - d

From $t = 6$ to $t = 12$ seconds, he walks towards the detector at a rate of $\frac{1}{3}$ feet per second. The distance function for $6\lt t\leq12$ is $d(t)=10-\frac{1}{3}(t - 6)=12-\frac{1}{3}t$. At $t = 12$ seconds, $d(12)=12-\frac{1}{3}\times12 = 8$ feet.

To find the distance from the detector when he finishes:
We use the last - part of the motion. When he finishes at $t = 12$ seconds, using the formula $d(t)=12-\frac{1}{3}t$ (for $6\lt t\leq12$), substituting $t = 12$ gives $d = 8$ feet.

Answer:

8 feet