Question

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 far will the snail be from the rock after 9 minutes?

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 \(v=\frac{\text{distance}}{\text{time}}\), so \(v = \frac{1}{\frac{4}{3}}=\frac{3}{4}\) inches per minute.

Step2: Calculate distance after 9 minutes

Using the formula \(d = v\times t\), where \(t = 9\) minutes and \(v=\frac{3}{4}\) inches per minute. So \(d=\frac{3}{4}\times9=\frac{27}{4} = 6\frac{3}{4}\) inches.

Answer:

\(6\frac{3}{4}\) inches (or \(\frac{27}{4}\) inches)