MIA-20260531-091D9E76

**Step‑by‑step reasoning** 1. **Define the situation** * Train A (Chicago → St. Louis) leaves Chicago at 10:00 AM, speed = **60 mph**. * Train B (St. Louis → Chicago) leaves St. Louis at 11:00 AM, speed = **80 mph**. * Distance between the cities = **300 mi** (south‑north line). We’ll work in *hours* after **10:00 AM**. 2. **How far is Train A when Train B starts?** Train A travels for 1 hour before Train B departs: \[ d_A(1\text{ h}) = 60\ \text{mph} \times 1\ \text{h}=60\ \text{mi} \] So at 11:00 AM the gap between the two trains is \[ 300\ \text{mi} - 60\ \text{mi}=240\ \text{mi}. \] 3. **Relative speed after 11:00 AM** Both trains are now moving toward each other, so their *relative* speed is the sum of their individual speeds: \[ v_{\text{rel}} = 60\ \text{mph} + 80\ \text{mph}=140\ \text{mph}. \] 4. **Time needed to close the 240‑mile gap** \[ t_{\text{close}} = \frac{\text{gap}}{v_{\text{rel}}} = \frac{240\ \text{mi}}{140\ \text{mph}} = \frac{12}{7}\ \text{h} = 1.7142857\ \text{h}. \] Convert the fractional hour to minutes and seconds: \[ 0.7142857\ \text{h}\times60\ \frac{\text{min}}{\text{h}} = 42.8571\ \text{min} \] \[ 0.8571\ \text{min}\times60\ \frac{\text{s}}{\text{min}} \approx 51.43\ \text{s}. \] So the closing phase lasts **1 hour 42 minutes 51 seconds**. 5. **Add this to the starting time of the second train** * Second train departs at **11:00 AM**. * Add the closing time: \[ 11{:}00\ \text{AM} + 1\ \text{h}\;42\ \text{min}\;51\ \text{s} = 12{:}42\ \text{min}\;51\ \text{s}\ \text{PM}. \] (If you round to the nearest minute, this is **12:43 PM**.) --- ### Verification with an algebraic approach Let * \(t\) = time after **10:00 AM** (in hours) when the trains meet. - Position of Train A (southward from Chicago): \[ d_A(t) = 60t. \] - Train B starts one hour later, so for \(t\ge 1\) its position (northward from St. Louis) is \[ d_B(t) = 300 - 80(t-1). \] Set the positions equal when they meet: \[ 60t = 300 - 80(t-1) \;\Longrightarrow\; 60t = 300 - 80t + 80 \;\Longrightarrow\; 140t = 380 \;\Longrightarrow\; t = \frac{380}{140}=2.7142857\text{ h}. \] **\(t=2.7142857\) h after 10:00 AM** equals \(2\) h \(+0.7142857\) h = 2 h 42 min 51 s → **12:42:51 PM**. Both methods give the same answer. --- ## Answer The two trains meet **approximately at 12:42 PM (more precisely 12 h 42 min 51 s after midnight, i.e., 12:42:51 PM).**