Car Overtaking Truck: Relative Speed & Catch-Up Problem

Distance, Rate & Time 9th-10th Grade
PROBLEM
A truck enters a highway driving 60 mph. A car enters the highway at the same place 13 minutes later and drives 67 mph in the same direction. From the time the car enters the highway, how long will it take the car to pass the truck?

Visualizing the Situation

Before diving into the algebra, let's picture what's happening on this highway.

A truck enters a highway driving 60 mph. A car enters the highway at the same place 13 minutes later and drives 67...

The truck gets a 13-minute head start before the car even enters. Once both vehicles are on the highway, the car needs to make up that initial gap while both vehicles continue moving forward.

What This Problem Teaches

  • Converting between time units (minutes to hours) for consistent calculations
  • Understanding relative speed as the rate at which one object gains on another
  • Setting up distance equations when objects start at different times
  • Recognizing that "catch-up" problems involve closing an initial gap
  • Interpreting fractional answers in practical time contexts

Solution: Method 1 — The Gap-Closing Approach

The most intuitive way to think about this problem is that the car needs to close the gap created by the truck's 13-minute head start.

Step 1 — Convert the head start time to hours

Since our speeds are in mph, we need consistent time units. Convert 13 minutes to hours:

13 minutes = 13/60 hours = 13/60 hours

Step 2 — Calculate the truck's head start distance

While the car is still off the highway, the truck travels for 13 minutes at 60 mph:

Head start distance = Rate × Time
Head start distance = 60 mph × (13/60) hours = 13 miles

Step 3 — Find the relative closing speed

Once the car enters the highway, both vehicles move in the same direction. The car closes the gap at the difference between their speeds:

Relative speed = Car speed - Truck speed
Relative speed = 67 mph - 60 mph = 7 mph

Step 4 — Calculate time to close the gap

Now we can find how long it takes the car to close the 13-mile gap at a relative speed of 7 mph:

Time = Distance ÷ Speed
Time = 13 miles ÷ 7 mph = 13/7 hours

Step 5 — Convert to mixed time format

To express this in a more practical format:

13/7 hours = 1.857... hours
= 1 hour + (6/7 × 60) minutes
= 1 hour + 51.43 minutes
≈ 1 hour 51 minutes

Solution: Method 2 — Equating Distances from Start

An alternative approach is to set up equations based on how far each vehicle has traveled from the original entry point.

Step 1 — Define the variable

Let t = hours the car drives until it catches the truck

Step 2 — Express the truck's total travel time

The truck has been driving for 13 minutes longer than the car:

Truck's driving time = t + 13/60 hours

Step 3 — Set up distance equations

When the car catches the truck, both vehicles are the same distance from the entry point:

Car's distance = Truck's distance
67t = 60(t + 13/60)
67t = 60t + 60 × (13/60)
67t = 60t + 13

Step 4 — Solve for t

Isolate the variable to find the catch-up time:

67t = 60t + 13
67t - 60t = 13
7t = 13
t = 13/7 hours

This confirms our answer from Method 1: the car will catch the truck after driving for 13/7 hours.

The car will catch and pass the truck 13/7 hours after the car enters the highway, which is approximately 1 hour and 51 minutes.

Verification

Let's check our answer by calculating where both vehicles are when the car catches the truck.

Car's position after 13/7 hours:

Distance = 67 mph × (13/7) hours = 67 × 13/7 = 871/7 ≈ 124.43 miles

Truck's position after 13/7 hours:
Remember, the truck has been driving for 13/7 + 13/60 hours total.

Total truck time = 13/7 + 13/60 = 780/420 + 91/420 = 871/420 hours
Distance = 60 mph × (871/420) hours = 60 × 871/420 = 52260/420 ≈ 124.43 miles

✓ Both vehicles are at the same position, confirming our answer is correct.

Common Pitfalls

✗ Mistake 1: Adding the speeds instead of subtracting
Some students calculate closing speed as 67 + 60 = 127 mph. This would be correct if the vehicles were moving toward each other, but they're moving in the same direction. The car only gains on the truck at the difference: 67 - 60 = 7 mph.
✗ Mistake 2: Forgetting to convert time units
Using 13 minutes directly with speeds in mph leads to unit mismatches. Always convert to consistent units first: 13 minutes = 13/60 hours before calculating with mph.
✗ Mistake 3: Calculating from when the truck started
The question asks "from the time the car enters the highway." Some students add the 13-minute head start to their answer, but the clock starts ticking when the car begins driving, not when the truck started.
✗ Mistake 4: Assuming the car immediately catches up
A few students think the car catches the truck in 13 minutes because that's the head start time. But during those 13 minutes, the truck continues moving forward, so it takes longer to close the gap.

The Underlying Pattern

This problem follows the standard relative motion formula for same-direction pursuit:

Catch-up time = Head start distance ÷ Relative speed

Where:
• Head start distance = (slower speed) × (head start time)
• Relative speed = (faster speed) - (slower speed)

This pattern appears whenever a faster object pursues a slower object that had a head start. The key insight is that once both objects are moving, the gap closes at the difference in their speeds, not their individual speeds.

Note: This formula only works when both objects maintain constant speeds and move in straight lines. Real-world complications like acceleration, traffic, or route changes would require more complex analysis.

Real Applications

  • Military and law enforcement: Calculating intercept times for pursuing vehicles or aircraft
  • Logistics and delivery: Determining when a faster truck can overtake a slower convoy
  • Sports analytics: Analyzing when a trailing runner can catch the leader in distance races
  • Network communications: Computing when faster data packets catch up to slower ones in network traffic analysis

What If?

1
Change the Car's Speed
A truck enters a highway driving 60 mph. A car enters the highway at the same place 13 minutes later and drives 72 mph in the same direction. From the time the car enters the highway, how long will it take the car to pass the truck?
Step 1 — Find the truck's head start distance

Head start distance = 60 mph × (13/60) hours = 13 miles

Step 2 — Calculate relative closing speed

Relative speed = 72 - 60 = 12 mph

Step 3 — Find catch-up time

Time = 13 miles ÷ 12 mph = 13/12 hours ≈ 1.08 hours

Step 4 — Convert to minutes

13/12 hours × 60 = 65 minutes = 1 hour 5 minutes

Answer: 13/12 hours, or about 1 hour 5 minutes

Verification: Car position: 72 × (13/12) = 78 miles. Truck position: 60 × (13/12 + 13/60) = 78 miles ✓

2
Reverse the Unknown
A truck enters a highway driving 60 mph. A car enters at the same place later and drives 67 mph in the same direction. The car catches the truck exactly 2 hours after the car enters the highway. How many minutes was the truck's head start?
Step 1 — Set up the distance equation

When they meet: Car distance = Truck distance
67 × 2 = 60 × (2 + head start in hours)

Step 2 — Solve for head start time

134 = 120 + 60 × (head start)
14 = 60 × (head start)
Head start = 14/60 hours

Step 3 — Convert to minutes

(14/60) hours × 60 minutes/hour = 14 minutes

Verification

Head start distance = 60 × (14/60) = 14 miles
Relative speed = 67 - 60 = 7 mph
Catch-up time = 14 ÷ 7 = 2 hours ✓

Answer: 14 minutes

3
Add a Constraint
A truck enters a highway driving 60 mph. A car enters the highway at the same place 13 minutes later and drives 67 mph. The car needs to catch the truck before reaching a weigh station that is 150 miles from the entry point. Will the car catch the truck in time?
Step 1 — Find when they meet

From the original problem: they meet after 13/7 hours ≈ 1.857 hours

Step 2 — Find meeting distance from entry point

Distance = Car's speed × time = 67 × (13/7) = 871/7 ≈ 124.4 miles

Step 3 — Compare to weigh station distance

Meeting point: 124.4 miles
Weigh station: 150 miles
124.4 < 150, so yes!

Step 4 — Calculate safety margin

Safety margin = 150 - 124.4 = 25.6 miles

Answer: Yes, the car will catch the truck about 25.6 miles before the weigh station

4
Three-Vehicle Chain
A truck (60 mph) enters a highway. A car (67 mph) enters 13 minutes later. A motorcycle (80 mph) enters 10 minutes after the car. How long after the motorcycle enters will it catch the car?
Step 1 — Find car's head start over motorcycle

Car drives alone for 10 minutes = 10/60 = 1/6 hour
Head start distance = 67 × (1/6) = 67/6 miles

Step 2 — Find relative speed

Motorcycle vs car relative speed = 80 - 67 = 13 mph

Step 3 — Calculate catch-up time

Time = distance ÷ relative speed
Time = (67/6) ÷ 13 = 67/78 hours

Step 4 — Convert to minutes

(67/78) hours × 60 = 4020/78 ≈ 51.5 minutes

Answer: 67/78 hours, or about 51.5 minutes after the motorcycle enters

Verification: Motorcycle travels 80 × (67/78) ≈ 68.7 miles. Car travels 67 × (67/78 + 1/6) ≈ 68.7 miles ✓

Frequently Asked Questions

How do you solve relative speed problems when one vehicle has a head start?+
First calculate the head start distance (rate × time for the lead vehicle), then divide by the difference in speeds. In this problem, the truck gets a 13-mile head start (60 mph × 13/60 hours), and the car closes the gap at 7 mph faster, taking 13/7 ≈ 1.86 hours to catch up.
What's the difference between relative speed and individual speeds in chase problems?+
Individual speeds tell you how fast each vehicle moves. Relative speed is how fast the gap between them closes - it's the difference between their speeds. Here, the car gains on the truck at 67 - 60 = 7 mph, even though both vehicles are moving much faster than 7 mph.
Why don't you add the speeds in same-direction motion problems?+
You only add speeds when objects move toward each other (opposite directions). When moving in the same direction, the faster object closes the gap at the difference of their speeds. If both move at 60 mph, the gap never closes - you need the speed difference.
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.

2026-08-19