Average Rate of Change - Math 1

1. A particle's position along a straight line is recorded at regular intervals. From $t = 2$ to $t = 8$ seconds, the particle moved from position $15$ meters to position $-9$ meters. However, at $t = 5$ seconds, the particle was at position $12$ meters. What can be concluded about the particle's motion?
   Answer: The average velocity from $t = 2$ to $t = 8$ was $-4$ m/s, but the particle changed direction at least once during this interval The average velocity from $t = 2$ to $t = 8$ is $\frac{-9 - 15}{8 - 2} = \frac{-24}{6} = -4$ m/s. However, since the particle was at position 12 at $t = 5$ (between the starting position 15 and some point in its journey), and ended at -9, the particle couldn't have moved in a straight line from 15 to -9. It must have changed direction. Option A incorrectly assumes consistent direction. Option C miscalculates the rate for the sub-interval. Option D confuses distance with displacement.
2. A population study tracked bacterial growth in three different samples. Sample A grew from $200$ to $800$ bacteria in $4$ hours. Sample B grew from $150$ to $600$ bacteria in $3$ hours. Sample C grew from $100$ to $500$ bacteria in $2$ hours. Which statement correctly compares their average growth rates?
   Answer: Sample C had the highest rate at $200$ bacteria/hour, followed by Sample B at $150$ bacteria/hour, then Sample A at $150$ bacteria/hour Calculate each rate: Sample A: $\frac{800-200}{4} = \frac{600}{4} = 150$ bacteria/hour. Sample B: $\frac{600-150}{3} = \frac{450}{3} = 150$ bacteria/hour. Sample C: $\frac{500-100}{2} = \frac{400}{2} = 200$ bacteria/hour. Sample C has the highest rate at 200, while A and B tie at 150. The multiplication factor is irrelevant to average rate of change.
3. Two functions $f(x)$ and $g(x)$ are defined on the interval $[0, 6]$. Given that $f(0) = 5$, $f(6) = 17$, $g(0) = 8$, and $g(6) = 20$, which statement about their average rates of change is correct?
   Answer: Function $f(x)$ has average rate $2$ and $g(x)$ has average rate $2$, but they may have different instantaneous rates throughout the interval For $f(x)$: average rate $= \frac{17-5}{6-0} = \frac{12}{6} = 2$. For $g(x)$: average rate $= \frac{20-8}{6-0} = \frac{12}{6} = 2$. Both have the same average rate of change, but this doesn't mean they increase at the same rate throughout the interval or that they're identical functions. They could have very different behaviors between the endpoints.
4. A stock's value changed from $\$25$ per share on Monday to $\$31$ per share on Friday. On Wednesday, it was worth $\$22$ per share. If someone claims the stock 'gained $\$1.50$ per day on average,' what is wrong with this reasoning?
   Answer: The calculation is correct, but average rate of change only considers endpoints, not intermediate values like Wednesday The average rate of change from Monday to Friday is $\frac{31-25}{5-1} = \frac{6}{4} = 1.50$ per day. This calculation is mathematically correct. However, average rate of change only depends on the starting and ending points of an interval, not on intermediate values. Wednesday's value of $22 doesn't affect the average rate calculation, though it shows the stock didn't increase uniformly.

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