Chase Problem: Catching Up at Different Speeds

Distance, Rate & Time 9th-10th Grade
Problem
Tanya, who is a long distance runner, runs at an average speed of 8 miles per hour. Two hours after Tanya leaves your house, you leave in your car and follow the same route. If your average speed is 40mph how long will it be before you catch up with Tanya? How far will you be from your house?

What This Problem Teaches

  • How to handle staggered start times in distance-rate-time problems
  • Setting up equations where two moving objects travel equal distances
  • Converting between total elapsed time and individual travel times
  • Visualizing pursuit scenarios with head starts
  • Checking answers using multiple calculation approaches
Tanya, who is a long distance runner, runs at an average speed of 8 miles per hour. Two hours after Tanya leaves your...

Solution: Method 1 — Equal Distance at Catch-Up

The key insight is that when you catch up to Tanya, both of you will have traveled the same total distance from your house.

Step 1 — Find Tanya's head start

Tanya runs for 2 hours before you start driving:

Head start distance = 8 mph × 2 hours = 16 miles

Step 2 — Set up the catch-up scenario

Let t = time (in hours) that you drive before catching up. When you catch up:

  • You will have driven for t hours at 40 mph
  • Tanya will have been running for (2 + t) hours total at 8 mph
  • Both distances from the house must be equal

Step 3 — Write the equal distance equation

Your distance = Tanya's distance
40t = 8(2 + t)

Step 4 — Solve for catch-up time

40t = 8(2 + t)
40t = 16 + 8t
40t - 8t = 16
32t = 16
t = 0.5 hours

Step 5 — Calculate the distance from house

Using your travel: Distance = 40 mph × 0.5 hours = 20 miles

You will catch up to Tanya after driving for 0.5 hours (30 minutes), and you will be 20 miles from your house.

Solution: Method 2 — Relative Speed Approach

Think of this as a gap-closing problem. Tanya has a head start, and you're closing the gap at a certain rate.

Step 1 — Calculate the initial gap

When you start driving, Tanya is already 16 miles ahead (as calculated above).

Step 2 — Find the relative speed

You're gaining on Tanya at the difference in your speeds:

Relative speed = 40 mph - 8 mph = 32 mph

This means the gap closes at 32 miles per hour.

Step 3 — Calculate time to close the gap

Time = Distance ÷ Speed
Time to catch up = 16 miles ÷ 32 mph = 0.5 hours

Step 4 — Find distance from house

In 0.5 hours, you drive: 40 × 0.5 = 20 miles

Same result: 0.5 hours (30 minutes) to catch up, 20 miles from house.

Verification

Let's check our answer by verifying both people have traveled the same distance from the house:

Tanya's total distance:
She runs for 2.5 hours total: 8 mph × 2.5 hours = 20 miles ✓

Your total distance:
You drive for 0.5 hours: 40 mph × 0.5 hours = 20 miles ✓

Both distances equal 20 miles, confirming our answer is correct.

Watch Out For These

✗ Forgetting the head start
Setting up 40t = 8t gives t = 0, which ignores that Tanya had a 2-hour head start. Always account for the time difference in pursuit problems.
✗ Using the wrong time for Tanya
Writing 40t = 8t instead of 40t = 8(2 + t). Remember: when you've been driving for t hours, Tanya has been running for 2 + t hours total.
✗ Calculating distance from the wrong starting point
Some students calculate how far Tanya traveled during the chase (4 miles) instead of her total distance from the house (20 miles). The question asks for distance from your house, not distance traveled during the pursuit.

The Pattern Behind This

Chase problems follow a general structure where you set equal the distances traveled by both moving objects. The key formula for staggered start problems is:

(Faster speed) × (Catch-up time) = (Slower speed) × (Head start + Catch-up time)

This pattern works whenever:

  • Both objects travel along the same path
  • One object has a head start in time
  • Speeds remain constant
  • You want to find when they're at the same location
When the faster object can't catch up: If the "faster" object is actually slower or equal in speed, there's no solution. Always check that the relative speed is positive before solving.

How to Spot This Problem Type

Look for these key phrases that signal a chase/pursuit problem:

  • "catches up" or "overtakes"
  • "leaves X hours later" or "head start"
  • "follows the same route" or "same path"
  • "when will they meet" in the context of pursuit

These problems differ from "meeting in the middle" scenarios because one person starts before the other, rather than both starting simultaneously from different locations.

Why This Matters

Chase problems model many real situations:

  • Emergency response: An ambulance catching up to a moving accident
  • Navigation: A faster ship catching a slower one that left port earlier
  • Sports: A cyclist making up time against a competitor with a head start
  • Technology: Network packets or data streams with different transmission delays

Try These Variations

1Different Head Start
Sarah runs at 6 mph and has a 3-hour head start. You drive at 45 mph on the same route. How long will you drive before catching up, and how far from the starting point will you be?
Step 1 — Calculate Sarah's head start distance

Distance = 6 mph × 3 hours = 18 miles

Step 2 — Set up the equal distance equation

Let t = your driving time. When you catch up:
Your distance = Sarah's distance
45t = 6(3 + t)

Step 3 — Solve for catch-up time

45t = 18 + 6t
39t = 18
t = 18/39 = 6/13 hours ≈ 0.46 hours

Step 4 — Calculate distance from start

Distance = 45 × (6/13) = 270/13 ≈ 20.8 miles

Step 5 — Verification

Sarah's total: 6 × (3 + 6/13) = 6 × 45/13 = 270/13 miles ✓

Answer: You'll catch up in about 28 minutes, 20.8 miles from the start.

2Adding a Constraint
Tanya runs at 8 mph with a 2-hour head start. You drive at 40 mph, but after 20 minutes of driving, you stop for 10 minutes to get gas. How long total (from when you first started driving) before you catch Tanya?
Step 1 — Calculate positions after gas stop

After 20 minutes (1/3 hour) driving + 10 minutes stopped:
• You've driven 1/3 hour, traveled 40 × 1/3 = 13.33 miles
• Tanya has run for 2 + 1/3 + 1/6 = 2.5 hours total
• Tanya's distance: 8 × 2.5 = 20 miles

Step 2 — Set up equation for remaining chase

When you resume, Tanya is 20 - 13.33 = 6.67 miles ahead.
Let t = additional driving time to catch up:
40t = 6.67 + 8t

Step 3 — Solve for additional time

32t = 6.67
t = 6.67/32 ≈ 0.208 hours = 12.5 minutes

Step 4 — Calculate total time

Total time from first start: 20 + 10 + 12.5 = 42.5 minutes

Step 5 — Verification

Your total distance: 13.33 + 40(0.208) = 21.67 miles
Tanya's distance: 8 × 2.708 = 21.67 miles ✓

Answer: 42.5 minutes total from when you first started.

3Three-Way Chase
Tanya runs at 8 mph. Two hours later, you drive at 40 mph. One hour after you start driving, your friend starts on a motorcycle at 60 mph. Who catches Tanya first, and when?
Step 1 — Find when you catch Tanya

From the original problem: You catch Tanya after 0.5 hours of driving.
This happens 2.5 hours after Tanya started.

Step 2 — Determine friend's position at your catch-up time

When you catch Tanya (2.5 hours after Tanya started), your friend has been driving for:
2.5 - 3 = -0.5 hours
This is negative, so your friend hasn't started yet!

Step 3 — Calculate when friend catches Tanya

Friend starts 3 hours after Tanya. Let t = friend's driving time:
60t = 8(3 + t)
60t = 24 + 8t
52t = 24
t = 24/52 = 6/13 ≈ 0.46 hours

Step 4 — Compare catch-up times

• You: 0.5 hours after you start = 2.5 hours after Tanya starts
• Friend: 0.46 hours after friend starts = 3.46 hours after Tanya starts

Step 5 — Verification

At 2.5 hours: Tanya at 20 miles, you at 20 miles ✓
At 3.46 hours: Tanya at 27.7 miles, friend at 27.7 miles ✓

Answer: You catch Tanya first, after 30 minutes of driving.

4Reverse Engineering
Tanya runs at 8 mph with a 2-hour head start. You want to catch her in exactly 45 minutes of driving. What speed must you maintain?
Step 1 — Determine Tanya's position when caught

Tanya will have run for: 2 + 0.75 = 2.75 hours
Her distance: 8 × 2.75 = 22 miles

Step 2 — Set up equation for required speed

You must travel 22 miles in 0.75 hours.
Let v = your required speed:
v × 0.75 = 22

Step 3 — Solve for required speed

v = 22 ÷ 0.75 = 22 × (4/3) = 88/3 ≈ 29.33 mph

Step 4 — Verification using relative speed

Gap to close: 16 miles
Relative speed: 29.33 - 8 = 21.33 mph
Time to close: 16 ÷ 21.33 = 0.75 hours ✓

Step 5 — Check reasonableness

29.33 mph is slower than the original 40 mph, which makes sense since you want to take longer (45 min vs 30 min) to catch up.

Answer: You must drive at 29.33 mph (or 88/3 mph exactly).

Frequently Asked Questions

Set up an equation where the distances traveled by both people are equal at the catch-up moment. If the faster person starts later, multiply their speed by their actual travel time. In this example, when the car catches up after driving for 0.5 hours, both will have traveled 20 miles from the house.
Total time counts from when the first person starts. Travel time counts from when each individual starts moving. Here, if Tanya runs for 2.5 total hours, the car only drives for 0.5 hours because it started 2 hours later.
The relative speed must be positive. Calculate how fast the gap closes by subtracting speeds: 40 mph - 8 mph = 32 mph closing speed. Since this is positive, the car will eventually catch up. If the speeds were equal or the slower person was faster, there would be no catch-up.
NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

Contact

This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-09-06

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