Question

1. sketch the graph, then answer the question below.

a. cooper starts 2 feet away from the motion detector.
b. he then walks away from the motion detector at a rate of 1 foot per second for 2 seconds.
c. he remains still for 4 seconds.
d. he then walks toward the detector at a rate of -\frac{1}{2} ft/sec for 4 seconds.
how far is cooper from the motion detector when he finishes her walk?
cooper is ______ ft away from the motion detector when he finishes.

Explanation:

Step1: Analyze each part of the motion

• Part a: Cooper starts \(2\) feet away. So at \(t = 0\), \(y=2\).
• Part b: He walks away at a rate of \(1\) foot per second for \(2\) seconds. The distance formula is \(y=y_0+vt\). Here \(y_0 = 2\), \(v = 1\), and \(t\) ranges from \(0\) to \(2\). So \(y=2 + 1\times t\) for \(0\leq t\leq2\). At \(t = 2\), \(y=2+2=4\).
• Part c: He remains still for \(4\) seconds. The distance does not change. So \(y = 4\) for \(2\lt t\leq6\).
• Part d: He walks toward the detector at a rate of \(-\frac{1}{2}\) ft/sec for \(4\) seconds. The distance formula is \(y=y_0+vt\). Here \(y_0 = 4\), \(v=-\frac{1}{2}\), and \(t\) ranges from \(6\) to \(10\). So \(y=4-\frac{1}{2}(t - 6)\). When \(t = 10\), \(y=4-\frac{1}{2}(10 - 6)=4 - 2=2\).

Step2: Calculate the final distance

We found that when he finishes (at \(t = 10\) seconds), using the formula for part d \(y=4-\frac{1}{2}(t - 6)\) with \(t = 10\), we get \(y = 2\) feet.

Answer:

\(2\)