d = 16t. The given equation represents the distance d, in inches, where t represents the number of seconds since...

1. TRANSLATE the equation components

• Given equation: \(\mathrm{d = 16t}\)
• \(\mathrm{d}\) represents distance in inches
• \(\mathrm{t}\) represents time in seconds since the object started moving
• \(\mathrm{16}\) is the coefficient multiplying the time variable

2. INFER what the coefficient represents

• In linear motion, when distance is proportional to time (\(\mathrm{d = vt}\)), the coefficient represents velocity
• The coefficient \(\mathrm{16}\) tells us how much the distance changes for each unit change in time
• Since distance increases by \(\mathrm{16}\) inches every \(\mathrm{1}\) second, this is the rate of motion

3. INFER the units and meaning

• \(\mathrm{Rate = \frac{change\:in\:distance}{change\:in\:time}}\)
• \(\mathrm{Rate = \frac{16\:inches}{1\:second} = 16\:inches\:per\:second}\)
• This confirms the object moves at a constant velocity of \(\mathrm{16}\) inches per second

Answer: C. The object is moving at a rate of 16 inches per second.

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students confuse the coefficient with total distance rather than rate.

Students might think "16 means the object moved 16 inches total" without considering that the equation shows distance depends on time. They focus on the number \(\mathrm{16}\) in isolation rather than understanding its role as a rate coefficient.

This may lead them to select Choice A (The object moved a total of 16 inches).

Second Most Common Error:

Poor TRANSLATE reasoning: Students misidentify what part of the equation represents the total distance.

Students might recognize that distance depends on time but incorrectly think "\(\mathrm{16t}\) represents the rate" instead of recognizing that \(\mathrm{16t}\) represents the total distance and \(\mathrm{16}\) alone represents the rate.

This may lead them to select Choice B (The object moved a total of 16t inches).

The Bottom Line:

This problem tests whether students understand the difference between a rate (coefficient) and a quantity (the entire expression). The key insight is recognizing that in \(\mathrm{d = 16t}\), the coefficient \(\mathrm{16}\) describes how fast distance changes, not how much distance has changed.