Question

problem 1
a snail is moving away from a rock at a constant rate. this table shows the distance the snail is from the rock at certain times.

distance (in.)	time (min)
1	$1\frac{1}{3}$

how many minutes does it take for the snail to reach a distance of 9 inches from the rock?

Explanation:

Step1: Find the speed of the snail

The snail moves 1 inch in \(1\frac{1}{3}\) minutes (which is \(\frac{4}{3}\) minutes). Speed \(v\) is distance over time, so \(v=\frac{1}{\frac{4}{3}}=\frac{3}{4}\) inches per minute? Wait, no, wait. Wait, actually, when distance is 1 inch, time is \(1\frac{1}{3}\) minutes. Wait, no, maybe I got it reversed. Wait, the formula for constant speed is \(d = vt\), where \(d\) is distance, \(v\) is speed, \(t\) is time. From the table, when \(d = 1\) inch, \(t=1\frac{1}{3}=\frac{4}{3}\) minutes. So to find \(v\), we can rearrange \(d=vt\) to \(v=\frac{d}{t}\). Wait, no, if we want to find time for a given distance, we can find the rate first. Wait, the snail's speed: when distance is 1 inch, time is \(\frac{4}{3}\) minutes? Wait, no, maybe the rate is distance per minute? Wait, no, let's think again. Let's find the time per inch. Wait, the problem is to find the time when distance is 9 inches. So first, find the time it takes to move 1 inch, then multiply by 9. From the table, when distance is 1 inch, time is \(1\frac{1}{3}=\frac{4}{3}\) minutes? Wait, no, wait the table: distance (in) is 0 at time 0, distance 1 at time \(1\frac{1}{3}\) minutes. So the time to move 1 inch is \(\frac{4}{3}\) minutes? Wait, no, that would mean the speed is \(\frac{1}{\frac{4}{3}}=\frac{3}{4}\) inches per minute. But we need time for 9 inches. So using \(t=\frac{d}{v}\), where \(v=\frac{3}{4}\) inches per minute. So \(t=\frac{9}{\frac{3}{4}}=9\times\frac{4}{3}=12\) minutes. Wait, let's check. Alternatively, since the relationship is linear (constant rate), we can set up a proportion. Let \(t\) be the time for 9 inches. Then \(\frac{1}{1\frac{1}{3}}=\frac{9}{t}\). Cross-multiplying: \(1\times t = 9\times1\frac{1}{3}\). \(1\frac{1}{3}=\frac{4}{3}\), so \(t = 9\times\frac{4}{3}=12\) minutes. Yes, that makes sense. So the proportion is \(\frac{\text{distance}_1}{\text{time}_1}=\frac{\text{distance}_2}{\text{time}_2}\), but since it's constant rate, \(\frac{d_1}{t_1}=\frac{d_2}{t_2}\), but actually, since \(d = vt\), then \(\frac{d_1}{t_1}=v\) and \(\frac{d_2}{t_2}=v\), so \(\frac{d_1}{t_1}=\frac{d_2}{t_2}\). Wait, no, if we have \(d_1 = 1\), \(t_1=\frac{4}{3}\), \(d_2=9\), \(t_2=t\), then \(\frac{1}{\frac{4}{3}}=\frac{9}{t}\)? Wait, no, that would be \(\frac{d_1}{t_1}=\frac{d_2}{t_2}\) implies \(t_2=\frac{d_2\times t_1}{d_1}\). So \(t_2=\frac{9\times\frac{4}{3}}{1}=12\). Yes, that works. So the time is 12 minutes.

Step1: Set up the proportion

Let \(t\) be the time (in minutes) for 9 inches. Since the snail moves at a constant rate, the ratio of distance to time is constant. So:

$$ \frac{1}{1\frac{1}{3}} = \frac{9}{t} $$

Step2: Convert mixed number to improper fraction

\(1\frac{1}{3}=\frac{4}{3}\), so the proportion becomes:

$$ \frac{1}{\frac{4}{3}} = \frac{9}{t} $$

Step3: Cross - multiply to solve for \(t\)

Cross - multiplying gives:

$$ 1\times t=9\times\frac{4}{3} $$

Step4: Simplify the right - hand side

\(9\times\frac{4}{3}=\frac{9\times4}{3}=3\times4 = 12\)
So \(t = 12\).

Answer:

12