MIA-20260531-D552634C

**Assumption:** Since the second train is heading north from St. Louis (toward Chicago), the first train must be heading south from Chicago (toward St. Louis) in order for them to meet. --- ### Step 1: Calculate the first train’s head start The first train travels alone from 10:00 AM to 11:00 AM (1 hour). \[ \text{Distance} = 60 \text{ mph} \times 1 \text{ hr} = 60 \text{ miles} \] At **11:00 AM**, Train 1 is **60 miles south of Chicago**, while Train 2 is just leaving St. Louis, **300 miles south of Chicago**. ### Step 2: Find the remaining distance between them at 11:00 AM \[ 300 \text{ miles} - 60 \text{ miles} = 240 \text{ miles} \] ### Step 3: Determine their combined closing speed The trains are now moving toward each other: - Train 1 southbound: **60 mph** - Train 2 northbound: **80 mph** \[ \text{Relative speed} = 60 + 80 = 140 \text{ mph} \] ### Step 4: Calculate the time needed to meet after 11:00 AM \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{140} = \frac{12}{7} \text{ hours} \] Convert \(\frac{12}{7}\) hours to hours and minutes: \[ \frac{12}{7} = 1\frac{5}{7} \text{ hours} = 1 \text{ hour} + \frac{5}{7}\times 60 \text{ minutes} \] \[ \frac{5}{7}\times 60 = \frac{300}{7} = 42\frac{6}{7} \text{ minutes} \] So the trains meet **\(1\) hour and \(42\frac{6}{7}\) minutes after 11:00 AM**. ### Step 5: Add the time to 11:00 AM \[ 11:00 \text{ AM} + 1 \text{ hr } 42\frac{6}{7} \text{ min} = \mathbf{12:42\frac{6}{7} \text{ PM}} \] --- ### Algebraic Check Let \(t\) = hours after **11:00 AM** that they meet. - Train 1’s distance south of Chicago: \(60 + 60t\) - Train 2’s distance south of Chicago: \(300 - 80t\) Set positions equal: \[ 60 + 60t = 300 - 80t \] \[ 140t = 240 \] \[ t = \frac{12}{7} \text{ hours} \quad \checkmark \] --- **Answer:** The trains meet at **12:42 and \(\frac{6}{7}\) minutes PM** (approximately **12:42:51 PM**).