a train leaves station M and travels west at 35 km/hr. one hour later a faster train leaves station M and also travels west on a parallel track at 40 km/hr. How far from the station M will the faster train catch the slower one?
There are two ways to do it, without algebra, and with algebrs:
Method 1; Without algebra:
You can do it this way in you head. When the faster train starts an
hour after the slower train left, the slower train has 35 km head start.
The faster train's catch-up rate is the difference in their speeds, or
40-35 or 5 km/hr. So to catch up the entire 35 km at 5 km/hr will take
7 hrs.
Method 2: With algebra.
Make this chart:
Distance Rate Time
-------------------------------------------
Slower train
Faster train
Put in the speeds:
Distance Rate Time
-------------------------------------------
Slower train 35
Faster train 40
Let t = the time it takes the faster train to catch
the slower train:
Distance Rate Time
-------------------------------------------
Slower train 35
Faster train 40 t
Then since the slower train travels 1 hour more than
the faster train, we put t+1 for the time of the slower
train:
Distance Rate Time
-------------------------------------------
Slower train 35 t+1
Faster train 40 t
Use Distance = Rate × time to indicate the multiplication
of the rate times the time.
Distance Rate Time
-------------------------------------------
Slower train 35(t+1) 35 t+1
Faster train 40t 40 t
Both trains had traveled the same distances when
the faster one catches the slower one. So we
set the two distances equal:
35(t+1) = 40t
35t + 35 = 40t
35 = 5t
7 = t
Answer: 7 hours.
Edwin