Worked solutions: Same-Direction Pursuit

All 8 problems from "Same-Direction Pursuit" with complete step-by-step worked solutions: recognising the type, the idea, the working steps and the conclusion. Use this page to review with your child after the animated lesson, or let them check their own working against it.

Every problem, fully solved

A car chases a bicycle travelling in the same direction. The car goes at 50 km/h and the bicycle at 20 km/h. By how many kilometres does the gap between them shrink each hour?

Key idea: Same direction: the gap closes at the DIFFERENCE of the speeds.

Recognise

A car chases a bicycle in the same direction: the car at 50 km/h, the bicycle at 20 km/h.

The idea

Both move forward, so the gap only shrinks by how much faster the car is.

Steps

1
In one hour the car covers 50 km while the bicycle covers 20 km.
2
The gap shrinks by 50 − 20 = 30 km each hour.

So: So the gap shrinks by 30 kilometres each hour.

A motorcycle travels at 15 km/h. A car behind it travels at 45 km/h in the same direction. How fast is the car gaining on the motorcycle?

60 km/h
30 km/h
45 km/h
15 km/h

Key idea: Same direction: subtract the speeds. Adding is only for head-on meetings.

Recognise

A car at 45 km/h follows a motorcycle at 15 km/h in the same direction.

The idea

The gaining speed is the difference of the two speeds.

Steps

1
Gaining speed: 45 − 15 = 30 km/h.
2
Not 60 km/h: that sum would only be right if they moved toward each other.

So: So the car is gaining at 30 km/h.

A delivery bike leaves the depot at 10 km/h. How many kilometres ahead is it after 2 hours?

Key idea: Head start distance = the leader's speed × its time alone.

Recognise

A delivery bike leaves the depot at 10 km/h — how far ahead is it after 2 hours?

The idea

Distance = speed × time, as usual.

Steps

1
The bike covers 10 km every hour.
2
10 × 2 = 20 km.

So: So it is 20 kilometres ahead.

A jogger runs at 8 km/h and has a 6 km head start. A cyclist follows at 14 km/h. In how many hours does the cyclist catch the jogger?

Key idea: Catch-up time = gap ÷ gaining speed.

Recognise

A jogger runs at 8 km/h with a 6 km head start; a cyclist follows at 14 km/h.

The idea

The cyclist closes the gap at the difference of the speeds.

Steps

1
Gaining speed: 14 − 8 = 6 km/h.
2
Time: 6 ÷ 6 = 1 hour.

So: So the cyclist catches the jogger in 1 hour.

A motorcycle leaves a town at 7:00 travelling at 24 km/h. A car leaves the same town at 8:00 travelling at 48 km/h in the same direction. How many hours after the car starts does it catch the motorcycle?

Key idea: Catch-up time = gap ÷ gaining speed — build the gap from the head start first.

Recognise

A motorcycle leaves at 7:00 at 24 km/h; a car leaves the same town at 8:00 at 48 km/h, same direction.

The idea

The motorcycle's one-hour head start sets the gap; the car closes it at the difference of speeds.

Steps

1
Head start: 24 × 1 = 24 km.
2
Gaining speed: 48 − 24 = 24 km/h.
3
Time: 24 ÷ 24 = 1 hour.

So: So the car catches the motorcycle 1 hour after the car starts.

A car travelling at 55 km/h chases a truck travelling at 40 km/h. The truck is 30 km ahead. The car catches the truck after ___ hours, at ___ km from the car's starting point.

Key idea: One chase, two questions: time first (gap ÷ gaining speed), then the chaser's distance.

Recognise

A car at 55 km/h chases a truck at 40 km/h that is 30 km ahead — the time and the catch point are both asked for.

The idea

Time from the gap and gaining speed; then the catch point is the car's own speed × that time.

Steps

1
Gaining speed: 55 − 40 = 15 km/h.
2
Time: 30 ÷ 15 = 2 hours.
3
Catch point: 55 × 2 = 110 km.
4
Check: the truck reaches 30 + 40 × 2 = 110 km too.

So: So the car catches the truck after 2 hours, at 110 km from the car's starting point.

A boat leaves a harbour at 9 km/h. Two hours later, a speedboat leaves the same harbour at 27 km/h in the same direction. How many kilometres from the harbour does the speedboat catch the boat?

Key idea: When the DISTANCE is asked for, finish with the chaser's speed × catch-up time.

Recognise

A boat leaves a harbour at 9 km/h; two hours later a speedboat follows at 27 km/h — the catch distance from the harbour is asked for.

The idea

Gap from the head start, time from the gaining speed, then distance with the speedboat's own speed.

Steps

1
Head start: 9 × 2 = 18 km.
2
Gaining speed: 27 − 9 = 18 km/h, so catch-up time = 18 ÷ 18 = 1 hour.
3
Catch point: 27 × 1 = 27 km from the harbour.

So: So the speedboat catches the boat 27 kilometres from the harbour.

A car starts 24 km behind a van. The car travels at 52 km/h and the van at 40 km/h, in the same direction. Which calculation gives the catch-up time?

24 ÷ (52 + 40)
24 ÷ (52 − 40)
24 × (52 − 40)
(52 − 40) ÷ 24

Key idea: Catch-up time = gap ÷ (faster − slower) — subtraction, because they move the same way.

Recognise

A car starts 24 km behind a van; the car travels at 52 km/h and the van at 40 km/h, same direction — pick the calculation.

The idea

The formula mirrors the method: the gap divided by the gaining speed.

Steps

1
Gaining speed: 52 − 40 = 12 km/h.
2
Time: 24 ÷ (52 − 40) = 24 ÷ 12 = 2 hours.

So: So the calculation is 24 ÷ (52 − 40).

Learn it with the animated lesson

The problem

A motorcycle leaves 0 km from warehouse AA at 07:00, traveling at 12 km/h. A car starts from 0 km at 08:00, traveling in the same direction at 36 km/h. When does the car catch the motorcycle, and how far from AA do they meet?

Progress

NaN%

Shortcuts: Space (play), Arrow keys (step), R (reset), F (fullscreen)

Problem overview

The motorcycle starts first at 07:00

Car: 36 km/h · Motorcycle: 12 km/h

The motorcycle leaves warehouse AA at 07:00, traveling at 12 km/h. The car does not leave until 08:00, at 36 km/h in the same direction. The car is much faster — but how long will it take to catch up?

Key idea

Same direction: the car cannot "eat" all 36 km/h — only the part that is faster.

Solution steps

0 / 6 steps

Try Your Own Numbers

Sandbox

Change the numbers and watch the diagram redraw itself:

Car speed (km/h)

Motorcycle speed (km/h)

Car departure time (hours)

Motorcycle departure time (hours)

Car start position from AA (km)

Motorcycle start position from AA (km)

Practise this type Same-Direction Pursuit

Every problem, fully solved

A car chases a bicycle travelling in the same direction. The car goes at 50 km/h and the bicycle at 20 km/h. By how many kilometres does the gap between them shrink each hour?

Key idea: Same direction: the gap closes at the DIFFERENCE of the speeds.

Recognise

A car chases a bicycle in the same direction: the car at 50 km/h, the bicycle at 20 km/h.

The idea

Both move forward, so the gap only shrinks by how much faster the car is.

Steps

1
In one hour the car covers 50 km while the bicycle covers 20 km.
2
The gap shrinks by 50 − 20 = 30 km each hour.

So: So the gap shrinks by 30 kilometres each hour.

A motorcycle travels at 15 km/h. A car behind it travels at 45 km/h in the same direction. How fast is the car gaining on the motorcycle?

60 km/h
30 km/h
45 km/h
15 km/h

Key idea: Same direction: subtract the speeds. Adding is only for head-on meetings.

Recognise

A car at 45 km/h follows a motorcycle at 15 km/h in the same direction.

The idea

The gaining speed is the difference of the two speeds.

Steps

1
Gaining speed: 45 − 15 = 30 km/h.
2
Not 60 km/h: that sum would only be right if they moved toward each other.

So: So the car is gaining at 30 km/h.

A delivery bike leaves the depot at 10 km/h. How many kilometres ahead is it after 2 hours?

Key idea: Head start distance = the leader's speed × its time alone.

Recognise

A delivery bike leaves the depot at 10 km/h — how far ahead is it after 2 hours?

The idea

Distance = speed × time, as usual.

Steps

1
The bike covers 10 km every hour.
2
10 × 2 = 20 km.

So: So it is 20 kilometres ahead.

A jogger runs at 8 km/h and has a 6 km head start. A cyclist follows at 14 km/h. In how many hours does the cyclist catch the jogger?

Key idea: Catch-up time = gap ÷ gaining speed.

Recognise

A jogger runs at 8 km/h with a 6 km head start; a cyclist follows at 14 km/h.

The idea

The cyclist closes the gap at the difference of the speeds.

Steps

1
Gaining speed: 14 − 8 = 6 km/h.
2
Time: 6 ÷ 6 = 1 hour.

So: So the cyclist catches the jogger in 1 hour.

Key idea: Catch-up time = gap ÷ gaining speed — build the gap from the head start first.

Recognise

A motorcycle leaves at 7:00 at 24 km/h; a car leaves the same town at 8:00 at 48 km/h, same direction.

The idea

The motorcycle's one-hour head start sets the gap; the car closes it at the difference of speeds.

Steps

1
Head start: 24 × 1 = 24 km.
2
Gaining speed: 48 − 24 = 24 km/h.
3
Time: 24 ÷ 24 = 1 hour.

So: So the car catches the motorcycle 1 hour after the car starts.

A car travelling at 55 km/h chases a truck travelling at 40 km/h. The truck is 30 km ahead. The car catches the truck after ___ hours, at ___ km from the car's starting point.

Key idea: One chase, two questions: time first (gap ÷ gaining speed), then the chaser's distance.

Recognise

A car at 55 km/h chases a truck at 40 km/h that is 30 km ahead — the time and the catch point are both asked for.

The idea

Time from the gap and gaining speed; then the catch point is the car's own speed × that time.

Steps

1
Gaining speed: 55 − 40 = 15 km/h.
2
Time: 30 ÷ 15 = 2 hours.
3
Catch point: 55 × 2 = 110 km.
4
Check: the truck reaches 30 + 40 × 2 = 110 km too.

So: So the car catches the truck after 2 hours, at 110 km from the car's starting point.

A boat leaves a harbour at 9 km/h. Two hours later, a speedboat leaves the same harbour at 27 km/h in the same direction. How many kilometres from the harbour does the speedboat catch the boat?

Key idea: When the DISTANCE is asked for, finish with the chaser's speed × catch-up time.

Recognise

A boat leaves a harbour at 9 km/h; two hours later a speedboat follows at 27 km/h — the catch distance from the harbour is asked for.

The idea

Gap from the head start, time from the gaining speed, then distance with the speedboat's own speed.

Steps

1
Head start: 9 × 2 = 18 km.
2
Gaining speed: 27 − 9 = 18 km/h, so catch-up time = 18 ÷ 18 = 1 hour.
3
Catch point: 27 × 1 = 27 km from the harbour.

So: So the speedboat catches the boat 27 kilometres from the harbour.

A car starts 24 km behind a van. The car travels at 52 km/h and the van at 40 km/h, in the same direction. Which calculation gives the catch-up time?

24 ÷ (52 + 40)
24 ÷ (52 − 40)
24 × (52 − 40)
(52 − 40) ÷ 24

Key idea: Catch-up time = gap ÷ (faster − slower) — subtraction, because they move the same way.

Recognise

A car starts 24 km behind a van; the car travels at 52 km/h and the van at 40 km/h, same direction — pick the calculation.

The idea

The formula mirrors the method: the gap divided by the gaining speed.

Steps

1
Gaining speed: 52 − 40 = 12 km/h.
2
Time: 24 ÷ (52 − 40) = 24 ÷ 12 = 2 hours.

So: So the calculation is 24 ÷ (52 − 40).

Learn it with the animated lesson

The problem

Progress

NaN%

Shortcuts: Space (play), Arrow keys (step), R (reset), F (fullscreen)

Problem overview

The motorcycle starts first at 07:00

Car: 36 km/h · Motorcycle: 12 km/h

Key idea

Same direction: the car cannot "eat" all 36 km/h — only the part that is faster.

Solution steps

0 / 6 steps

Try Your Own Numbers

Sandbox

Change the numbers and watch the diagram redraw itself:

Car speed (km/h)

Motorcycle speed (km/h)

Car departure time (hours)

Motorcycle departure time (hours)

Car start position from AA (km)

Motorcycle start position from AA (km)