Question: A train left a station at 2:00 p.m. and traveled 45 km/hr. A second train left the station at 4:00 p.m. and traveled 65 km/hr in the same direction. At what time will the second train catch the first?
1. **State the problem:**
Two trains start from the same station but at different times and speeds. The first train leaves at 2:00 p.m. traveling at 45 km/hr. The second train leaves at 4:00 p.m. traveling at 65 km/hr. We need to find the time when the second train catches up to the first.
2. **Define variables and formula:**
Let $t$ be the time in hours after 4:00 p.m. when the second train catches the first.
Distance traveled by first train by 4:00 p.m. is:
$$45 \times 2 = 90 \text{ km}$$
Distance traveled by first train after 4:00 p.m. is:
$$45t \text{ km}$$
Distance traveled by second train after 4:00 p.m. is:
$$65t \text{ km}$$
The second train catches the first when their distances from the station are equal:
$$90 + 45t = 65t$$
3. **Solve the equation:**
$$90 + 45t = 65t$$
Subtract $45t$ from both sides:
$$90 + \cancel{45t} = 65t - \cancel{45t}$$
$$90 = 20t$$
Divide both sides by 20:
$$\frac{90}{20} = \frac{20t}{20}$$
$$4.5 = t$$
4. **Interpret the result:**
$t = 4.5$ hours after 4:00 p.m. means:
$$4:00 \text{ p.m.} + 4.5 \text{ hours} = 8:30 \text{ p.m.}$$
**Final answer:** The second train will catch the first train at 8:30 p.m.