Motion problems · البرمجة مع مارتن

Distance, speed, time and equations: how to calculate two cars meeting and catching up with a friend on a bike.

When will they meet?

Two cars set off towards each other. A cyclist is catching up with a friend. A train leaves Prague, another leaves Brno. When will they meet? These problems have a reputation as a scare — and yet they stand on one formula you already know how to turn: s = v · t.

Distance = speed times time. Anyone who goes 60 km/h for two hours has travelled 120 km. That is all the physics you need.

The rest is translation into an equation, exactly like last time. Two situations are key:

They go towards each other: their distances add. They meet when together they have covered the whole distance between them.

One is catching the other: catching up happens when both have the same distance from the start.

As x you almost always mark time — because from setting off it runs the same for both. Draw the situation as a line, substitute into s = v · t for each one separately and the equation almost builds itself.

Filip set off by bike to the cottage at 12 km/h. After an hour dad finds Filip forgot the keys and sets off after him by car at 60 km/h. Ema (sitting next to dad) calculates: “Filip has a 12 km head start. Every hour we close by 60 − 12 = 48 km.” Dad: “So 12 : 48 = a quarter of an hour.” In 15 minutes they really see Filip by the woods. Ema writes into her phone: head start divided by the difference of speeds. “An equation would say the same: 60t = 12 + 12t.”

💡

Always draw a line: who sets off from where, in which direction and at what speed. A picture in ten seconds replaces ten minutes of confusion.

Two patterns: towards each other and catching up

Pattern 1: towards each other. The towns are 210 km apart. A car from the first goes 60 km/h, from the second 80 km/h, they set off at the same time. When will they meet?

Mark: t = travel time (hours) — the same for both.

Distances: first car 60t, second 80t. When they meet they have together the whole 210 km:

60t + 80t = 210 → 140t = 210 → t = 1.5.

They meet in 1.5 hours. Check: 90 + 120 = 210 km. It fits.

Pattern 2: catching up. A walker sets off at 4 km/h. An hour later a cyclist sets off after them at 12 km/h. How long until they catch up?

Mark: t = the cyclist’s travel time. The walker has been going an hour longer: (t + 1).

Catching up = the same distance: 12t = 4(t + 1) → 12t = 4t + 4 → 8t = 4 → t = 0.5.

They catch up in half an hour. Check: cyclist 6 km, walker in 1.5 h also 6 km. It fits.

Watch units: km/h wants hours, not minutes. Write half an hour as 0.5 h.

Example 1: trains towards each other

From towns 360 km apart two trains set off towards each other, one 70 km/h, the other 50 km/h. When will they meet?

t = travel time of both trains.

Equation: 70t + 50t = 360 → 120t = 360 → t = 3 hours.

Check: the first covers 210 km, the second 150 km, together 360. It fits.

A helper for a fast head: they close at 70 + 50 = 120 km/h, so 360 : 120 = 3. The equation and plain sense say the same.

Example 2: catching up with a head start

Adam ran out to training at 8 km/h. Half an hour later mum set off after him by bike at 16 km/h. How long until she catches him?

t = mum’s travel time. Adam has been running 0.5 h longer: (t + 0.5).

The same distance: 16t = 8(t + 0.5) → 16t = 8t + 4 → 8t = 4 → t = 0.5 h.

She catches him in half an hour. Check: mum 8 km, Adam after an hour of running also 8 km. It fits.

Example 3: how far from town?

An extra question from the last example: how far from home does mum catch Adam?

I take the calculated t = 0.5 h and substitute into the distance of either of them:

Mum: s = 16 · 0.5 = 8 km.

I check with Adam: s = 8 · 1 = 8 km. The same number — that is a good sign, catching up means the same distance.

Motion problems often have two questions: WHEN (that is t from the equation) and WHERE (that is substituting t into s = v · t). Do not forget to read the question to the end.

⚠️

Do not mix minutes and hours. Speed in km/h + time in minutes = nonsense. Half an hour is 0.5 h, a quarter of an hour 0.25 h, 20 minutes is a third of an hour. Convert the time before you build the equation.

Practice: now you do it 🎲

Now try it for real. You will see all the questions at once, like on paper. البداية with ten. When it goes well, add more.

Practise equations — a motion problem is just an equation dressed up as cars and trains.

Practise the questions

On paper

A line, a table s, v, t for every traveller, an equation, an answer in a sentence.

The towns are 300 km apart. Cars set off towards each other at 40 km/h and 60 km/h. Draw a line, build the equation and calculate how long until they meet and where.
Tereza is walking on a trip at 5 km/h. An hour later Jonáš sets off on a scooter at 15 km/h. How long until he catches her and how many kilometres does he ride?
A cyclist rides 3 hours at 18 km/h. What speed would they need to do the same route in 2 hours? (First calculate the distance.)
Invent your own catching-up problem with two people from your family and solve it.

Now you 💪

I can tell apart “towards each other” (I add the distances) and “catching up” (the distances are the same).
I choose time as x and I watch who travels how long.
My units match: km/h with hours.

✅

Done when: You have three solved problems with sketches and answers in a sentence, one you invented yourself, and a run of five correct in practice.

What to take from this lesson

All motion stands on s = v · t.
Towards each other: the distances add up to the whole distance.
Catching up: the distances are equal; watch a head start in time.
x is almost always time. Drawing a line saves confusion.
WHEN = t from the equation, WHERE = substituting t into s = v · t.