Consider the formula that relates distance, velocity, and time (d = vt). If you are interested in the time it takes for a mouse traveling at a speed of 5 feet per minute to run across a 30-yard field, which unit conversion makes the most sense to use?

Question

Converting yards to feet by multiplying 30 yards by 1 yard / 3 feet
Do you think this makes sense?

Answer 1

The formula you provided, \( d = vt \), relates distance (d), velocity (v), and time (t). To solve for time, you would rearrange the formula to \( t = \frac{d}{v} \).

In your scenario, you have a distance of 30 yards and a velocity of 5 feet per minute. Since the velocity is given in feet per minute, it makes sense to convert the distance from yards to feet so that both quantities are in the same unit.

To convert yards to feet, you should multiply the number of yards by 3 feet per yard, not by \( 1 \text{ yard} / 3 \text{ feet} \). So, you would do:

\[ 30 \text{ yards} \times 3 \text{ feet/yard} = 90 \text{ feet} \]

After converting the distance, you can use the distance in feet (90 feet) along with the velocity in feet per minute (5 feet/minute) to find the time:

\[ t = \frac{d}{v} = \frac{90 \text{ feet}}{5 \text{ feet/minute}} = 18 \text{ minutes} \]

In summary, converting yards to feet makes sense, but make sure to multiply by the correct conversion factor of 3 feet per yard.

Answer 2

So it would be multiply 30 by 3 feet / 1 yard.

Answer 3

Yes, exactly! When converting from yards to feet, you would multiply by the conversion factor \( \frac{3 \text{ feet}}{1 \text{ yard}} \). This is how you set it up:

\[ 30 \text{ yards} \times \frac{3 \text{ feet}}{1 \text{ yard}} = 90 \text{ feet} \]

This ensures that the "yards" unit cancels out, leaving you with the distance in feet. Then you can use this distance along with the speed to find the time taken to cross the field.