Draw both trains from the same station.
They start in the same place and stop in the same place.
Why the distances are equal
“Catch up” means side by side. Both trains are then the same number of miles from the station, so D₁ and D₂ match.
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Catching Up
One starts late and chases — catching up means equal distances.
The problem
A freight train leaves a station at 40 mph. Three hours later an express train leaves the same station in the same direction at 60 mph. How long will it take the express to catch up?
The freight is 3 × 40 = 120 miles ahead when the express starts. The express closes the gap at 60 − 40 = 20 mph, so it needs 120 ÷ 20 = 6 hours. Use the difference of the speeds — same direction means they subtract. D (9 hours) is the freight's total time, 3 + 6: the question times the express, which only starts at hour 3. B just repeats the 3-hour head start.
Turn “three hours later” into numbers.
Let t be the express train’s travelling time.
Freight time
t + 3
Express time
t
Plus 3, not minus 3
The freight has been rolling longer, so its time is the bigger one. Adding 3 to the late train instead is the classic slip.
Fill a speed × time = distance table.
Distance is always speed times that train’s own time.
| Train | Speed | Time | Distance |
|---|---|---|---|
| Freight | 40 | t + 3 | 40(t + 3) |
| Express | 60 | t | 60t |
Keep the bracket — the 40 multiplies the whole time.
Two clocks this time
In earlier problems both rows shared one t. Here the trains left at different moments, so the times differ by exactly 3.
Catching up means the distances are equal.
D₁ = D₂
40(t + 3) = 60t
Nothing is added or subtracted here. Two different trips ending at the same milepost are simply equal.
Solve for t.
40(t + 3) = 60t
from Step 4
40t + 120 = 60t
distribute the 40
120 = 20t
− 40t both sides
6 = t
÷ 20 both sides
Read the question again
t was the express train’s time, so the answer is 6 hours from its departure — 9 hours after the freight left.
If the question asks where, put t back into either row of the table: the express runs 60 × 6 = 360 miles, so they meet 360 miles from the station.
Variations: a head start in minutes, an unknown speed, no catch-up at all.
Same equation, D₁ = D₂; the unknown moves around.
Minutes and clock times: a van leaves at 7:00 a.m. at 48 km/h; a car follows at 7:40 a.m. at 64 km/h. When does the car catch it?
40 min = 4060 = 23 h
hours to match km/h
48(t + 23) = 64t
t is the car’s time
48t + 32 = 64t, so t = 2
32 km ÷ 16 km/h
7:40 + 2 h = 9:40 a.m.
count from the car’s start
Convert minutes before they meet a speed. Then add t to the chaser’s start time, not the van’s.
40 minutes is not 0.4 hours
48(t + 0.4) = 64t gives t = 1.2 h and 8:52 a.m. Divide minutes by 60: 40 min is 23 h.
The speed is the unknown: a bike at 18 km/h has a 1.5-hour head start, and a motorbike catches it 2 hours after setting off. How fast is the motorbike?
bike’s time: 1.5 + 2 = 3.5 h
both times are known
18 × 3.5 = 63 km
where they meet
r × 2 = 63, so r = 31.5 km/h
the motorbike covers the same 63 km
When both times are given, the meeting distance comes first and the speed follows from it. If the meeting point is given instead, each time is distance ÷ speed: caught 72 km out, a bike at 18 km/h has ridden 72 ÷ 18 = 4 h; a chaser who left 1.5 h later has ridden 2.5 h, so its speed is 72 ÷ 2.5 = 28.8 km/h.
No solution: a walker at 5 km/h has a 2-hour head start, and the chaser also walks at 5 km/h.
5(t + 2) = 5t
the usual equation
5t + 10 = 5t, so 10 = 0
false for every t
never catches up
the 10 km gap never closes
The chaser must be faster. A false equation like 10 = 0 is the algebra saying so; a slower chaser gives a negative t.
Chasing to a set lead.
The gap to close is the head start plus the lead wanted.
head start: 4 × ½ = 2 km
the walker’s lead when the runner sets off
closing speed: 12 − 4 = 8 km/h
the runner gains 8 km each hour
level after 2 ÷ 8 = ¼ h = 15 min
the catch-up alone
2 km ahead: (2 + 2) ÷ 8 = ½ h
close the gap, then open a new one
Check: in ½ h the runner covers 6 km; the walker has walked 1 h, 4 km. The runner is 2 km ahead ✓.
The lead is extra distance
Stopping at the catch-up time (15 min) answers a different question. Every extra kilometre of lead costs another ⅛ h at this closing speed.
On the exam
Catch-up items give a head start in time and ask when, where or how fast. The fastest route is head-start distance ÷ difference in speeds; then check that both travellers are the same distance from the start (or the required gap apart) at your answer.
Check: are both trains at the same milepost?
Freight
40 · 9 = 360 miles
it ran 6 + 3 = 9 hours
Express
60 · 6 = 360 miles
it ran 6 hours
Now you try
Practice problem
The problem
A cyclist leaves home at 12 kph. Two hours later a motorbike leaves the same place on the same road and catches the cyclist in 30 minutes. How fast is the motorbike?
The cyclist starts 2 × 12 = 24 km ahead and rides another 6 km during the half hour, so the catch happens 30 km from home. The motorbike covers that 30 km in 0.5 h, so it does 60 kph. C forgets the 6 km added during the chase.
Hint: both times are already known here — the missing number is a speed. Write 30 minutes as 0.5 hour before you start.