Draw the road first.
A picture tells you whether the distances add or subtract.
Why this picture matters
They start together and move apart, so the 300 km is made of two pieces sitting end to end. That is why the distances add.
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Distance Apart
Two vehicles, opposite directions — the gap grows from both ends.
The problem
Two vehicles leave the same point at the same time in opposite directions. Car 1 goes 40 kph and Car 2 goes 60 kph. How long until they are 300 km apart?
Moving apart, the speeds add: the gap grows at 40 + 60 = 100 kph, so 300 ÷ 100 = 3 hours. D uses only the faster car and A only the slower one — neither opens the gap on its own.
Fill a speed × time = distance table.
One row per vehicle. Distance is always the last column.
| Vehicle | Speed | Time | Distance |
|---|---|---|---|
| Car 1 | 40 | t | 40t |
| Car 2 | 60 | t | 60t |
distance = speed · time, so 40 · t and 60 · t
The step everyone rushes
Both times are the same letter t. They left together and you stop the clock on both at once, so there is only one unknown.
Add the two distances.
D₁ + D₂ = Dapart
40t + 60t = 300
Opposite directions means the two trips lie end to end. Same direction would mean you subtract instead.
Combine the two t terms.
40t + 60t = 300
40t and 60t are like terms — both count the same t.
What 100 actually means
40 kph + 60 kph = 100 kph
The gap between them grows 100 km every hour.
You can jump straight to 100t = 300 once you trust this. The table just proves where it comes from.
Solve for t.
100t = 300
from Step 4
t = 3
÷ 100 both sides
Check your units
km ÷ kph = hours
The question asked how long, so the answer must be a time. Ending with km would mean a wrong set-up.
Variations: unknown speeds, minutes, a right-angle split.
The sum of speeds still runs the problem; these change what we solve for, or break the rule.
The speeds are unknown: two trucks leave together in opposite directions, one 10 km/h faster. After 2 hours they are 260 km apart. Find both speeds.
2r + 2(r + 10) = 260
r is the slower speed
4r + 20 = 260, so r = 60
subtract 20, divide by 4
60 km/h and 70 km/h
check: 120 + 140 = 260 ✓
Time is known and speed is the letter. Give the speed the question names: the faster one is 70, not the r = 60 we solved for.
Time in minutes: two scooters leave together in opposite directions at 30 and 45 km/h. How far apart are they after 20 minutes?
20 min = 13 h
km/h needs hours
(30 + 45) × 13 = 75 ÷ 3 = 25 km
sum of speeds × time
75 × 20 = 1,500 km
Multiplying km/h by minutes gives an answer sixty times too big. Convert the minutes first.
Not opposite but at right angles: one car drives north at 30 km/h and another east at 40 km/h from the same corner. How far apart after 1.5 hours?
north 30 × 1.5 = 45 km; east 40 × 1.5 = 60 km
the two legs
gap = √45² + 60² = √2,025 + 3,600
Pythagoras
= √5,625 = 75 km
a 3-4-5 triangle, ×15
At right angles the distances do not add (45 + 60 = 105 is wrong); they form the legs of a right triangle.
A rest stop: give each traveller only the time they are moving. Two walkers leave together in opposite directions at 4 and 6 km/h, and the faster one rests for half an hour. After 2 hours the slower has walked 4 × 2 = 8 km and the faster 6 × 1.5 = 9 km, so they are 8 + 9 = 17 km apart.
A staggered start, in opposite directions.
Give the early one its head start, then let the gap grow at the combined speed.
by 8:00 the bus is 50 km away
1 h × 50 km/h
50 + (50 + 70)t = 290
the gap grows 120 km each hour
120t = 240, so t = 2
hours after 8:00
Check: the bus drove 3 h (150 km), the car 2 h (140 km), and 150 + 140 = 290 ✓. Moving toward each other is the same sum of speeds, shrinking the gap: cars 360 km apart at 50 and 70 km/h meet after 360 ÷ 120 = 3 hours.
t starts at the second departure
t = 2 counts from 8:00, so the answer is 10:00, not 9:00. Write down which clock t belongs to.
On the exam
Opposite-direction items — apart or toward each other — always use the sum of the speeds. Watch for a staggered start (give the early one its head start first) and for minutes that must become hours, then check that the two distances add to the total given.
Check: put 3 hours back into both trips.
Car 1
40 · 3 = 120 km
in 3 hours
Car 2
60 · 3 = 180 km
in 3 hours
Now you try
Practice problem
The problem
Two boats leave the same dock at the same time in opposite directions. One sails at 18 kph. After 4 hours they are 140 km apart. How fast is the other boat?
In 4 hours they open a 140 km gap, so together they manage 140 ÷ 4 = 35 kph. Take off the known boat: 35 − 18 = 17 kph. A is that combined rate, which is not either boat's speed.
Hint: the blank has moved. Here the time is known and a speed is missing, but the table and the adding stay the same.