10. Rate, Distance, and Time Problems

Introduction

This session models situations with two moving objects — such as two vehicles, or one person
walking toward another — as a linear system using the relationship distance equals rate times
time.

Definition

Using \(d=rt\), a rate-distance-time system sets up one linear equation per traveler
(distance as a function of time, possibly with a head start), then solves for the time or
position at which the two distances are equal.

Real-Life Uses

• Two trains leaving different stations at different times and speeds, and finding when they
  meet.
• Two hikers starting from opposite ends of a trail, walking toward each other.
• A cyclist chasing a runner who has a head start, to find the catch-up time.

Why You Should Know This

Rate-distance-time systems test the same linear-system skills as every other session, but
disguised inside physical motion — recognizing the underlying \(d=rt\) structure quickly is the
key translation skill this session builds.

Worked Examples

Example 1

Problem: Car A leaves a station traveling at 60 mph. Car B leaves the same station 1 hour later traveling at 75 mph in the same direction. When does Car B catch Car A?

Solution: Let \(t\) be hours after Car B leaves. Car A's distance: \(d=60(t+1)\). Car B's distance: \(d=75t\). Set equal: \(75t=60t+60\), so \(15t=60\) and \(t=4\). Car B catches Car A 4 hours after Car B departs.

Example 2

Problem: Two hikers start 21 miles apart and walk toward each other, one at 3 mph and the other at 4 mph. When do they meet?

Solution: Let \(t\) be hours until they meet. Combined distance covered equals the gap: \(3t+4t=21\), so \(7t=21\) and \(t=3\). They meet after 3 hours.

Example 3

Problem: A boat travels downstream 60 miles in the same time it travels upstream 36 miles. If the boat's speed in still water is \(b\) and the current speed is \(c\), set up the system.

Solution: Downstream speed is \(b+c\) and upstream speed is \(b-c\); since time is equal, \(\frac{60}{b+c}=\frac{36}{b-c}\), which rearranges to the linear system \(60(b-c)=36(b+c)\), or \(60b-60c=36b+36c\), giving \(24b=96c\), so \(b=4c\).

Example 4

Problem: Two trains 300 miles apart travel toward each other, one at 50 mph and one at 70 mph. How long until they meet?

Solution: Combined speed is \(50+70=120\) mph. Time to close a 300-mile gap: \(t=\frac{300}{120}=2.5\) hours.

Example 5

Problem: A cyclist rides at 18 mph and a runner at 6 mph starts 2 hours earlier from the same point in the same direction. When does the cyclist catch the runner?

Solution: Let \(t\) be hours after the cyclist starts. Runner's distance: \(d=6(t+2)\). Cyclist's distance: \(d=18t\). Set equal: \(18t=6t+12\), so \(12t=12\) and \(t=1\). The cyclist catches the runner 1 hour after starting.

Example 6

Problem: A student models a head-start problem as \(60t=75t+60\) instead of \(75t=60t+60\), reversing which racer has the head start. How does this change the setup, and why is it wrong here?

Solution: The equation must place the head-start term with the traveler who left later but moves faster catching up to the one with a lead; here Car A has the 1-hour head start, so its distance is \(60(t+1)\), not Car B's. Swapping the head start onto the wrong traveler produces a negative or nonsensical meeting time.