Example

Solving a Catch-Up Uniform Motion Problem Where One Driver Leaves 15 Minutes Later

Apply the seven-step problem-solving strategy for systems of linear equations to a uniform motion catch-up scenario where a fractional time delay is involved.

Problem: Charlie left his mother’s house traveling at an average speed of 3636 miles per hour. His sister Sally left 1515 minutes (14\frac{1}{4} hour) later traveling the same route at an average speed of 4242 miles per hour. How long before Sally catches up to Charlie?

  1. Read the problem and draw a diagram showing both drivers traveling the same route. Create a rate–time–distance table.
  2. Identify: The travel time for each driver.
  3. Name: Let cc = Charlie's driving time (in hours) and ss = Sally's driving time (in hours). Identify the rates: Charlie's rate is 3636 mph, Sally's rate is 4242 mph. Multiply rate by time to fill in the distance column:
Rate (mph)Time (hrs)Distance (miles)
Charlie3636cc36c36c
Sally4242ss42s42s
  1. Translate into a system of equations. Sally catches Charlie when they have traveled the same distance: 36c=42s36c = 42s. Since Sally left 14\frac{1}{4} hour later, her time is 14\frac{1}{4} hour less than Charlie's: s=c14s = c - \frac{1}{4}. The system is:
ight.$$ 5. Solve by substitution. Substitute $$s = c - \frac{1}{4}$$ into the second equation: $$36c = 42\left(c - \frac{1}{4} ight)$$ Distribute: $$36c = 42c - 10.5$$ (or $$36c = 42c - \frac{21}{2}$$). Subtract $$42c$$ from both sides: $$-6c = -10.5$$. Divide by $$-6$$: $$c = 1.75$$. Substitute $$c = 1.75$$ into the first equation: $$s = 1.75 - 0.25 = 1.5$$. 6. Check: Charlie: $$36 \cdot 1.75 = 63$$ miles. Sally: $$42 \cdot 1.5 = 63$$ miles. Both distances are equal. $$\checkmark$$ 7. Answer: It will take Sally $$1.5$$ hours (or $$1$$ hour and $$30$$ minutes) to catch up to Charlie.

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Updated 2026-04-24

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