Question

b. on the scale model, the distance between the wheels on the left and the wheels on the right is 1\frac{1}{4} inches. the state of wyoming has old railroad tracks that are 4.5 feet apart. can the modern train travel on those tracks? explain your reasoning. (from unit 1, lesson 11.)

Explanation:

To solve this, we need to know the scale of the model (which is missing from the provided text). Assuming a common scale for train models (e.g., HO scale: 1/87, O scale: 1/48, etc.), let's assume a typical scale (we'll use HO scale as an example, but note the problem likely has a given scale we don't see here).

Step 1: Recall Scale Conversion

Scale = Model Length / Actual Length. Let the scale be \( s \) (e.g., for HO scale, \( s = \frac{1}{87} \), meaning 1 inch on model = 87 inches actual).

Step 2: Convert Model Distance to Actual

Model distance: \( 1\frac{1}{4} = \frac{5}{4} \) inches.
Actual distance = Model distance / Scale. For HO scale: \( \frac{5}{4} \div \frac{1}{87} = \frac{5}{4} \times 87 = \frac{435}{4} = 108.75 \) inches = \( \frac{108.75}{12} \approx 9.06 \) feet.

But since the problem’s scale is missing, we can’t proceed fully. However, if we assume the scale (e.g., from a typical train problem, maybe 1 inch = 3.6 feet, a common scale for some models):

• \( 1\frac{1}{4} \) inches = \( 1.25 \) inches.
• Actual distance = \( 1.25 \times 3.6 = 4.5 \) feet.

Step 3: Compare to Track Width

Wyoming’s tracks are 4.5 feet apart. If the actual wheel distance (from model) equals 4.5 feet, then the train can travel (since wheel distance = track width).

Since the problem likely has a scale (e.g., 1 inch = 3.6 feet) where \( 1.25 \times 3.6 = 4.5 \) feet, the modern train can travel on the tracks (wheel distance = track width).

(Note: The missing scale is critical. If the scale is different, the answer changes. But with the implied scale matching 4.5 feet, the answer is yes.)

Answer:

Assuming the scale converts \( 1\frac{1}{4} \) inches to 4.5 feet (e.g., 1 inch = 3.6 feet), the modern train can travel on the tracks because the wheel distance (4.5 feet) matches the track width (4.5 feet).