Example

Solving a Catch-Up Uniform Motion Problem Where One Driver Leaves 1 Hour Later

Apply the seven-step problem-solving strategy for systems of linear equations to a uniform motion catch-up scenario, using two variables and substitution to solve the resulting system.

Problem: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 6060 miles per hour. Clark left Detroit 11 hour later traveling at a speed of 7575 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

  1. Read the problem and draw a diagram showing both drivers traveling the same route from Detroit to Orlando. Create a rate–time–distance table.
  2. Identify: The travel time for each driver.
  3. Name: Let mm = Mitchell's driving time (in hours) and cc = Clark's driving time (in hours). Identify the rates: Mitchell's rate is 6060 mph, Clark's rate is 7575 mph. Multiply rate by time to fill in the distance column:
Rate (mph)Time (hrs)Distance (miles)
Mitchell6060mm60m60m
Clark7575cc75c75c
  1. Translate into a system of equations. Clark catches Mitchell when they have traveled the same distance: 60m=75c60m = 75c. Since Clark left 11 hour later, his time is 11 hour less than Mitchell's: c=m1c = m - 1. The system is:
ight.$$ 5. Solve by substitution. Substitute $$c = m - 1$$ into the second equation: $$60m = 75(m - 1)$$ Distribute: $$60m = 75m - 75$$. Subtract $$75m$$ from both sides: $$-15m = -75$$. Divide by $$-15$$: $$m = 5$$. Substitute $$m = 5$$ into the first equation: $$c = 5 - 1 = 4$$. 6. Check: Mitchell: $$60 \cdot 5 = 300$$ miles. Clark: $$75 \cdot 4 = 300$$ miles. Both distances are equal. $$\checkmark$$ 7. Answer: It will take Clark $$4$$ hours to catch Mitchell.

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Updated 2026-04-24

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