MIA-20260531-276E2936

**Solution Overview** We have two cities, Chicago and St Louis, that are **300 mi** apart. - Train A leaves Chicago at **10 am**, heading **south** (toward St Louis) at **60 mph**. - Train B leaves St Louis at **11 am**, heading **north** (toward Chicago) at **80 mph**. We must find the clock time when the two trains meet. --- ## 1. Set up a time reference Let \[ t = \text{hours after 10:00 am} \] Thus - Train A starts at \(t=0\). - Train B starts at \(t=1\) hour (11 am). --- ## 2. Write the distance each train has traveled (in the north‑positive direction) If we call the northward direction **positive**: | Train | Start position | Velocity | Distance travelled (as a function of \(t\)) | |-------|----------------|----------|---------------------------------------------| | A (Chicago) | \(0\) mi (Chicago) | \(-60\) mph (south) | \(-60t\) | | B (St Louis) | \(-300\) mi (south of Chicago) | \(+80\) mph (north) | \(-300 + 80(t-1)\) for \(t\ge 1\) | *Why the “\(t-1\)” factor for Train B?* Train B does not start moving until 11 am, i.e. one hour after the reference time. --- ## 3. Equate positions – the meeting condition At the meeting moment the two trains are at the same coordinate: \[ -60t = -300 + 80(t-1), \qquad t\ge 1 \] --- ## 4. Solve for \(t\) \[ \begin{aligned} -60t &= -300 + 80t - 80 \\ -60t - 80t &= -300 - 80 \\ -140t &= -380 \\ t &= \frac{380}{140} = \frac{19}{7}\;\text{hours} \\ t &\approx 2.7142857\ \text{hours after 10:00 am} \end{aligned} \] --- ## 5. Convert the decimal hour to clock time \[ \frac{19}{7}\text{ h}=2\text{ h}+\frac{5}{7}\text{ h} \] \[ \frac{5}{7}\text{ h}= \frac{5}{7}\times 60\ \text{min}= \frac{300}{7}\ \text{min} = 42\ \text{min}+ \frac{6}{7}\ \text{min} \] \[ \frac{6}{7}\ \text{min}= \frac{6}{7}\times 60\ \text{s}= \frac{360}{7}\ \text{s} \approx 51.4\ \text{s} \] So the meeting occurs at \[ 10{:}00\ \text{am} + 2\ \text{h}\;42\ \text{min}\;51\ \text{s} \approx 12{:}42{:}51\ \text{pm}. \] Rounded to the nearest minute, **they meet at about 12:43 pm**. --- ## 6. Quick sanity‑check (relative‑speed method) 1. At 11 am (one hour after Train A starts) Train A has already covered \[ 60\ \text{mph}\times1\ \text{h}=60\ \text{mi}. \] Distance still between them: \(300-60=240\) mi. 2. After 11 am both trains are moving toward each other, so their **relative speed** is \[ 60\ \text{mph}+80\ \text{mph}=140\ \text{mph}. \] 3. Time needed to close the remaining 240 mi is \[ \frac{240\ \text{mi}}{140\ \text{mph}} = \frac{12}{7}\ \text{h} \approx 1.7142857\ \text{h}=1\ \text{h}\;42\ \text{min}\;51\ \text{s}. \] 4. Adding that to 11 am gives the same result: 12:42 pm (≈12:43 pm). Both approaches agree. --- ## 7. Final Answer **The trains meet at approximately 12 : 42 : 51 pm (about 12:43 pm).** --- *All reasoning steps have been shown, including the algebraic set‑up, the arithmetic, and a quick relative‑speed verification.*