Questions: The Caldwells are moving across the country. Mr. Caldwell leaves 3 hours before Mrs. Caldwell. if he averages 50 mph and she averages 90 mph, how many hours will it take Mrs. Caldwell to catch up to Mr. Caldwell?

The Caldwells are moving across the country. Mr. Caldwell leaves 3 hours before Mrs. Caldwell. if he averages 50 mph and she averages 90 mph, how many hours will it take Mrs. Caldwell to catch up to Mr. Caldwell?

Transcript text: The Caldwells are moving across the country. Mr. Caldwell leaves 3 hours before Mrs. Caldwell. if he averages 50 mph and she averages 90 mph, how many hours will it take Mrs. Caldwell to catch up to Mr. Caldwell?

Solution

Solution Steps

Step 1: Define the variables

Let \( t \) be the number of hours Mrs. Caldwell travels until she catches up to Mr. Caldwell.

Step 2: Express the distances traveled

Mr. Caldwell's distance: Since he leaves 3 hours earlier, his travel time is \( t + 3 \) hours. His distance is \( 50(t + 3) \) miles.
Mrs. Caldwell's distance: Her travel time is \( t \) hours. Her distance is \( 90t \) miles.

Step 3: Set the distances equal and solve for \( t \)

To catch up, the distances must be equal: \[ 50(t + 3) = 90t \] Expand and simplify: \[ 50t + 150 = 90t \] Subtract \( 50t \) from both sides: \[ 150 = 40t \] Divide both sides by 40: \[ t = \frac{150}{40} = 3.75 \]

Final Answer

\(\boxed{3.75}\) hours

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