How fast, how far, how long — and the three short rules that connect them.
CHAPTER 1.1 · RATE, DISTANCE, TIME
What "per" really means
The word per is small but it does most of the work in this whole
topic. Every time you see it, "per" means "for every".
"$5 per hour" = "for every hour, you get $5"
"60 km per hour" = "for every hour that passes, the car covers 60 km"
"3 cookies per child" = "for every child, 3 cookies"
A rate is exactly that — a "this for every that." When the rate
has the word time in the bottom (per hour, per minute, per second), we
usually call it a speed.
A rate is a recipe. "60 km/h" is the recipe: every hour, add another 60 km
to the total distance traveled. If you let the recipe run for 3 hours,
you've added 60 km three times — which is 180 km.
This is the only big idea in the whole chapter. Once you really feel that a
rate is a recipe that keeps being applied, every formula in this lesson is just a
way to count up how many times the recipe ran.
FEEL IT
Read each rate and finish the sentence in your head:
A
"The faucet drips at 2 drops per second" → in 10 seconds, you'd have ___ drops.
Answer
20 drops.
The recipe (2 drops) ran 10 times.
B
"Riya types at 40 words per minute" → in 6 minutes, she has typed ___ words.
Answer
240 words.
40 × 6.
CHAPTER 1.2 · RATE, DISTANCE, TIME
The triangle: D = R × T
Three quantities show up in every motion problem:
D — Distance (how far the traveler went)
R — Rate, also called speed (how fast)
T — Time (how long they travelled)
They are linked by one short rule:
D = R × T
"distance equals rate times time"
need D? D = R × T
need T? T = D ÷ R
need R? R = D ÷ T
If you know any two of the three, you can find the missing one. The next three
sub-chapters walk through each case in turn.
UNITS — THE THING MOST MISTAKES COME FROM
D = R × T only works when the units line up.
If R is in km per hour, then T must be in hours,
and you'll get D in kilometres.
If T is given in minutes but R is in km/h, convert first:
30 minutes = 0.5 hour.
The unit of D in your answer is "the top unit of R" — so km/h × hours = km.
CHAPTER 1.3 · RATE, DISTANCE, TIME
Finding the distance
You know how fast and how long. You want to know how far.
WORKED EXAMPLE
A train travels at a steady speed of 80 km/h for 4 hours.
How far does it travel?
The recipe: every hour, add 80 km. The recipe ran 4 times. So the total is
80 × 4 = 320 km.
The train travels 320 km.
MORE LIKE THIS
A
A cyclist rides at 18 km/h for 2 hours. How far?
Answer
18 × 2 = 36 km.
B — UNITS
A car drives at 50 km/h for 30 minutes. How far?
C — TWO LEGS
A bus drives at 70 km/h for 2 hours, then at 90 km/h for 3 hours. How far in total?
Answer
Leg 1: 70 × 2 = 140 km.
Leg 2: 90 × 3 = 270 km. Total = 410 km.
CHAPTER 1.4 · RATE, DISTANCE, TIME
Finding the time
Now you know how far and how fast. You want to know how long the trip takes.
WORKED EXAMPLE
A bus travels at 60 km/h. How long does it take to cover
240 km?
Rearrange the triangle: T = D ÷ R.
T = 240 ÷ 60 = 4 hours.
Think of it as "how many recipes did I have to run?" One recipe adds 60 km. To
reach 240 km, I need 4 recipes. Each recipe takes 1 hour. So 4 hours.
4 hours.
MORE LIKE THIS
A
An aeroplane flies at 800 km/h. How long does it take to fly 2,000 km?
Answer
2000 ÷ 800 = 2.5
hours (2 hours 30 minutes).
B
A walker covers 12 km at 4 km/h. How long does she walk?
Answer
12 ÷ 4 = 3 hours.
C — TWO PARTS
A driver covers 150 km at 50 km/h, then another 60 km at 30 km/h. How long
was the whole trip?
Answer
Leg 1: 150 ÷ 50 = 3 hours.
Leg 2: 60 ÷ 30 = 2 hours. Total = 5 hours.
CHAPTER 1.5 · RATE, DISTANCE, TIME
Finding the rate
You know how far and how long. You want to know the speed.
WORKED EXAMPLE
A runner covered 21 km in 3 hours. What was
her average speed?
Rearrange: R = D ÷ T = 21 ÷ 3 = 7 km/h.
7 km per hour.
"AVERAGE SPEED" — A SMALL TRAP
In real life the runner doesn't run at exactly 7 km/h the whole time — she
speeds up and slows down. The 7 km/h figure is her average over the
full trip. It's the constant speed that would have produced the same total
distance in the same total time.
For all the problems in this lesson, treat the given speeds as average speeds
unless the problem says otherwise.
MORE LIKE THIS
A
A train covers 360 km in 4 hours. What was its average speed?
Answer
360 ÷ 4 = 90 km/h.
B — UNITS
A snail crawls 30 cm in 5 minutes. What is its speed in cm/min?
Answer
30 ÷ 5 = 6 cm/min.
C — AVERAGE OVER TWO LEGS (TRICKY)
A car drives 60 km at 60 km/h, then another 60 km at 30 km/h. What is the
average speed for the whole trip?
Answer
Leg 1: 60 ÷ 60 = 1 hour.
Leg 2: 60 ÷ 30 = 2 hours. Total distance = 120 km. Total time = 3 hours.
Average = 120 ÷ 3 = 40 km/h.
It is NOT 45 km/h (the simple average of 60 and 30) — averages of
speeds over different times don't work that way. Always use total distance ÷
total time.
★ MINI-QUIZ · RATE BASICS
Try a few before moving on
Q1 · CH 1.3
A car drives at 40 km/h for 3 hours. How far does it go?
km
D = R × T = 40 × 3 = 120 km.
Q2 · CH 1.4
A walker covers 20 km at 4 km/h. How many hours does it take?
hours
T = D ÷ R = 20 ÷ 4 = 5 hours.
Q3 · CH 1.5
A train covered 400 km in 5 hours. What was its average
speed (km/h)?
km/h
R = D ÷ T = 400 ÷ 5 = 80 km/h.
CHAPTER 2.1 · COMING TOGETHER
The simple meeting
Now there are two travelers. They start at the same moment, at opposite
ends of a road, and walk towards each other. Where and when do they meet?
WORKED EXAMPLE
Two towns are 240 km apart. Anya drives east from her town at
50 km/h. Ben drives west from his town at 70 km/h.
They start at the same time. How long until they meet?
In every hour, Anya covers 50 km and Ben covers 70 km — at the same
time. So the gap between them shrinks by 50 + 70 = 120 km every hour.
Total gap to close: 240 km. Gap-closing rate: 120 km/h. Time =
240 ÷ 120 = 2 hours.
Anya travels 50 × 2 = 100 km. Ben travels 70 × 2 = 140 km.
Check: 100 + 140 = 240 km ✓ — they meet exactly when their two distances cover
the whole road.
They meet after 2 hours. Anya has covered 100 km, Ben 140 km.
COMING TOWARDS
When two travelers move towards each other, their
speeds add. The gap closes at R_A + R_B.
time to meet = gap ÷ (R_A + R_B)
MORE LIKE THIS
A
Two boats are 90 km apart and head towards each other at 12 km/h and 18 km/h.
When do they meet?
Answer
Combined = 30 km/h. Time =
90 ÷ 30 = 3 hours.
B — FIND ONE SPEED
Towns are 200 km apart. A car leaves the east town heading west at 60 km/h.
A truck leaves the west town at the same time heading east. They meet after
2 hours. How fast was the truck?
Answer
Total combined speed =
200 ÷ 2 = 100 km/h. Car was 60, so truck was 100 − 60 = 40 km/h.
CHAPTER 2.2 · COMING TOGETHER
When start times differ
Real life rarely starts at the same moment. One traveler leaves earlier and gets
a "head start" before the other one even begins. The trick is to pause the
clock at the moment the second one starts and ask: where is the first one
right now?
WORKED EXAMPLE
Towns are 300 km apart. Anya leaves the east town at
9 a.m., heading west at 40 km/h. Ben leaves
the west town at 10 a.m., heading east at 60 km/h.
At what time do they meet?
Step 1 — Where is Anya at 10 a.m.?
Anya travelled alone for 1 hour, covering 40 × 1 = 40 km. So at
10 a.m., the gap is no longer 300 km — it's 300 − 40 = 260 km.
Step 2 — From 10 a.m. onwards, they close the gap together
Step 3 — Add to the moment they're both travelling
10 a.m. + 2 h 36 min = 12:36 p.m.
They meet at 12:36 p.m.
STAGGERED STARTS
(1) Compute the head-start distance the early traveller covered alone.
(2) Subtract it from the original gap.
(3) Run the standard "meeting" formula on the new gap, starting from when the
second traveller began.
MORE LIKE THIS
A
Two cities are 500 km apart. A truck leaves City A at 8 a.m. driving east at
60 km/h. A car leaves City B at 9 a.m. driving west at 90 km/h. When do they
meet?
Answer
By 9 a.m. the truck has
travelled 60 km. New gap = 440 km. Combined speed from 9 a.m. = 150 km/h.
Time = 440 ÷ 150 ≈ 2.93 hr = 2 h 56 min. Meeting time = 9:00 + 2:56 =
11:56 a.m.
★ MINI-QUIZ · MEETING
Two coming-together problems
Q1 · CH 2.1
Two cars start 270 km apart and drive towards each other at
40 km/h and 50 km/h, starting at the same time. How many hours until they meet?
hours
Combined speed = 90 km/h. Time = 270 ÷ 90 =
3 hours.
Q2 · CH 2.2
Towns are 540 km apart. A leaves at 9 a.m. driving east at
60 km/h. B leaves at 10 a.m. driving west at 100 km/h. How many hours after B
starts do they meet?
hours
By 10 a.m., A has gone 60 × 1 = 60 km. New gap =
480 km. Combined speed from 10 a.m. = 160 km/h. Time = 480 ÷ 160 =
3 hours. (Meet at 1 p.m.)
CHAPTER 3.1 · CATCHING UP
Why the gap shrinks slowly
Now the two travelers are heading the same direction. The faster one
is behind, chasing the slower one. The question: how long does it take the
faster one to catch up?
Here's the key difference: when both move the same way, the slower one is
also moving away — so the gap shrinks only by the difference in speeds,
not the sum.
Every hour, the fast car gains 70 km of road. But the slow car ALSO covers 40 km
in that same hour — moving forward. So the actual distance closed
between them is only 70 − 40 = 30 km per hour.
That number — the difference in speeds — is called the
relative speed. It's the only thing you need.
SAME DIRECTION
When two travelers go the same way, their speeds subtract.
The gap shrinks at R_fast − R_slow.
time to catch up = head-start gap ÷ (R_fast − R_slow)
CHAPTER 3.2 · CATCHING UP
Head-start problems
The most common shape: someone leaves first, then someone else (faster) leaves
later and chases them.
WORKED EXAMPLE
A truck leaves at 9 a.m. driving east at 40 km/h. A car leaves
from the same place at 11 a.m., also heading east, at
80 km/h. When does the car catch the truck?
Step 1 — How big is the head start?
By 11 a.m., the truck has been driving for 2 hours. Head-start distance =
40 × 2 = 80 km. The car must close this 80 km gap.
Step 2 — Relative speed
Both moving east. Relative speed = 80 − 40 = 40 km/h.
Step 3 — Time to close
Time = 80 ÷ 40 = 2 hours. So the car catches the truck at
11:00 + 2:00 = 1:00 p.m.
The car catches up at 1:00 p.m.
Both have travelled the same distance from the start: car did 80 × 2 = 160 km,
truck did 40 × (2 + 2) = 160 km ✓.
MORE LIKE THIS
A
Anya starts walking at 4 km/h. One hour later, Ben starts cycling along the
same path at 12 km/h. How long does it take Ben to catch her?
Answer
Head start = 4 km.
Relative speed = 12 − 4 = 8 km/h. Time = 4 ÷ 8 = 0.5 hours
(30 minutes).
B
A bus left at 7 a.m. driving north at 50 km/h. A motorbike left at 9 a.m.
from the same place at 90 km/h. What time does the motorbike pass the bus?
Answer
Head start (by 9 a.m.) =
50 × 2 = 100 km. Relative speed = 40 km/h. Time = 100 ÷ 40 = 2.5 hr. Catches
up at 11:30 a.m.
CHAPTER 3.3 · CATCHING UP
Laps on a track
A circular track is a clever case of catch-up. Two runners start at the same
point and run the same way. The faster one will eventually lap the slower
one — that means they meet again, with the faster one one full lap ahead.
WORKED EXAMPLE
Two runners start at the same place and time, going the same way around a
400 m track. One runs at 5 m/s, the other at
8 m/s. After how many seconds does the faster runner lap the
slower one?
To "lap" means to be exactly 400 m (one full lap) ahead. So the gap to close
is 400 m. Relative speed = 8 − 5 = 3 m/s.
Time = 400 ÷ 3 ≈ 133.3 seconds.
About 133 seconds (2 minutes 13 seconds).
Laps are just catch-up problems where the head-start is the full track length.
Same formula, same idea — just dressed in running shoes.
MORE LIKE THIS
A
Two cyclists race around a 600 m circular track. Slow one: 6 m/s. Fast one:
9 m/s. When does the fast one first lap the slow one (in seconds)?
A walker leaves at 10 a.m. heading east at 5 km/h. A jogger
leaves the same point at 11 a.m. heading east at 10 km/h. How many hours after
the jogger starts does she catch the walker?
hours
Head start (at 11 a.m.) = 5 × 1 = 5 km. Relative
speed = 10 − 5 = 5 km/h. Time = 5 ÷ 5 = 1 hour.
Q2 · CH 3.3 (LAP)
Two runners on a 300 m track, same direction. Slow runner:
4 m/s. Fast runner: 7 m/s. After how many seconds does the fast runner first
lap the slow one?
seconds
Gap to close = 300 m. Relative = 3 m/s. Time =
300 ÷ 3 = 100 seconds.
CHAPTER 4.1 · UPSTREAM / DOWNSTREAM
What a current does
A river isn't just water — it's moving water. The current is a giant
conveyor belt carrying everything floating in it downstream. A boat with its
engine off would still drift downstream at the speed of the current.
Now turn the engine on. Let:
b = the boat's speed in still water (no current)
c = the speed of the current itself
Going downstream (with the current)
The boat's engine pushes it at b. The current pushes it at
c. They add up.
downstream speed = b + c
Going upstream (against the current)
The boat's engine pushes forward at b. The current pushes back
at c. They subtract.
upstream speed = b − c
Note: the boat's own engine speed (b) doesn't change. The current is
what makes the boat look faster or slower from the riverbank.
CHAPTER 4.2 · UPSTREAM / DOWNSTREAM
The add & subtract trick
The most common problem type: you're told the down-speed and the up-speed
(often via "covered X km in Y hours" each way). Find the boat's still-water speed
and the current.
WORKED EXAMPLE
A boat travels 60 km downstream in 3 hours, and the same
60 km upstream in 5 hours. Find the boat's speed in still water
and the speed of the current.
Compute each one-way speed first:
downstream speed = 60 ÷ 3 = 20 km/h → b + c = 20
upstream speed = 60 ÷ 5 = 12 km/h → b − c = 12
Add the two equations: (b + c) + (b − c) = 20 + 12.
The currents cancel — you get 2b = 32, so b = 16.
Subtract: (b + c) − (b − c) = 20 − 12. The boats
cancel — you get 2c = 8, so c = 4.
Boat: 16 km/h in still water. Current: 4 km/h.
Check: down = 16 + 4 = 20 ✓, up = 16 − 4 = 12 ✓.
BOAT & CURRENT
Given downstream speed D and upstream speed U:
boat in still water = (D + U) ÷ 2
current = (D − U) ÷ 2
CHAPTER 4.3 · UPSTREAM / DOWNSTREAM
Planes, escalators, and friends
The upstream/downstream pattern shows up everywhere there's a moving medium.
Whenever something moves through a thing that is itself moving, the same
add-and-subtract logic applies.
SAME FAMILY
Plane & wind: with the wind = engine + wind. Against
the wind = engine − wind. Most flight scheduling math uses exactly this.
Escalator: walking up a down-escalator = your walking
speed − escalator speed. Walking down a down-escalator = your speed +
escalator speed.
Conveyor belt: walking on a moving walkway at the
airport. Adds to your speed if you walk along; subtracts if against.
Treadmill: same idea — your running speed plus the
belt's reverse speed = your true ground-speed of zero (you stay in place).
WORKED EXAMPLE — PLANE
A plane flies 1,500 km with the wind in 3 hours, and the same
1,500 km against the wind in 5 hours. Find the plane's speed
in still air and the wind speed.
With-wind speed = 1500 ÷ 3 = 500 km/h.
Against-wind speed = 1500 ÷ 5 = 300 km/h.
Plane = (500 + 300) ÷ 2 = 400 km/h. Wind = (500 − 300) ÷ 2 =
100 km/h.
FEEL IT
ESCALATOR
Riya stands still on a down-escalator and rides it from top to bottom in
30 seconds. Walking on the still escalator takes her 60 seconds (it would
take 60 s on stairs that aren't moving). How long does it take her to walk
down a moving down-escalator?
Answer
Let escalator length = L.
Escalator speed = L / 30. Riya's walking speed = L / 60. Walking down a moving
down-escalator: combined speed = L/30 + L/60 = 3L/60 = L/20. Time = L ÷ (L/20)
= 20 seconds.
★ MINI-QUIZ · UPSTREAM/DOWNSTREAM
Two boat / wind problems
Q1 · CH 4.2 (BOAT)
A boat goes 18 km downstream in 1 hour and the same 18 km
upstream in 1.5 hours. What is the speed of the boat in still water (km/h)?
km/h
Downstream = 18 km/h. Upstream = 18/1.5 = 12 km/h.
Boat = (18 + 12) ÷ 2 = 15 km/h. (Current = 3 km/h.)
Q2 · CH 4.3 (PLANE)
A plane flies 600 km with the wind in 2 hours and the same
600 km against the wind in 3 hours. What is the wind speed (km/h)?
km/h
With wind = 300 km/h. Against = 200 km/h. Wind =
(300 − 200) ÷ 2 = 50 km/h. (Plane = 250 km/h.)
CHAPTER 5.1 · MASTERY
Common setup mistakes
If your answer doesn't check out, you almost certainly fell into one of these.
Skim the list and re-read your setup.
Mistake 1 — Mixed-up units
Multiplying km/h × minutes will give you nonsense. Convert times to hours (or
speeds to per-minute) before you compute. Most "off by a factor of 60" errors
come from this.
Mistake 2 — Added when you should have subtracted
Two travelers same direction: the gap closes at the
difference of their speeds, not the sum. If your "time to catch up" came
out tiny, you probably added.
Mistake 3 — Subtracted when you should have added
Two travelers opposite directions, meeting: speeds add. If
your "time to meet" came out huge, you probably subtracted.
Mistake 4 — Forgot the head start
In staggered-start problems, the early traveler is already somewhere when the
second one begins. Pause the clock at the second traveler's start time and
redo the gap.
Mistake 5 — Averaging speeds wrong
If a trip has two legs at different speeds, the average speed is NOT the simple
average of the two speeds. Use total distance ÷ total time.
Set up before you compute. Write down which traveller, which direction, which
speed. The arithmetic is easy once the setup is right.
★ FINAL TEST ★
Motion · Final Test
Ten problems mixing everything. Decide which chapter's idea applies, set up
before you compute, type your answer.
PROBLEM 1 · CH 1.3 (D = R × T)
A car drives at 50 km/h for 3 hours. How far does it go?
km
50 × 3 = 150 km.
PROBLEM 2 · CH 1.4 (T = D ÷ R)
A bus needs to cover 280 km at 70 km/h. How many hours
does the trip take?
hours
280 ÷ 70 = 4 hours.
PROBLEM 3 · CH 1.5 (R = D ÷ T)
A runner covered 480 m in 80 seconds. What was her average
speed in m/s? (careful — answer in m/s)
m/s
480 ÷ 80 = 6 m/s. (That's a healthy
jogging pace — about 22 km/h.)
PROBLEM 4 · CH 2.1 (MEETING)
Two trains start 320 km apart and head towards each other
at 70 km/h and 90 km/h, leaving at the same time. How many hours until they
meet?
hours
Combined = 160 km/h. Time = 320 ÷ 160 =
2 hours.
PROBLEM 5 · CH 2.1 (MEETING · FIND DISTANCE)
Two cars start 360 km apart and drive towards each other at
50 km/h and 70 km/h, at the same time. How far has the SLOWER car travelled
when they meet (km)?
km
Combined = 120 km/h. Time = 360 ÷ 120 = 3 hours.
Slower car: 50 × 3 = 150 km. (Faster car: 70 × 3 = 210 km.
Check: 150 + 210 = 360 ✓.)
PROBLEM 6 · CH 3.2 (CATCH-UP)
A truck leaves at 8 a.m. driving at 50 km/h. A car leaves
from the same place at 10 a.m. at 100 km/h. How many hours after the car
starts does it catch the truck?
hours
Head start (by 10 a.m.) = 50 × 2 = 100 km. Relative
speed = 100 − 50 = 50 km/h. Time = 100 ÷ 50 = 2 hours.
PROBLEM 7 · CH 3.3 (LAP)
Two runners on a 1,200 m circular track go the same way at
7 m/s and 12 m/s. After how many seconds does the faster runner first lap the
slower one?