Sheila leaves on a long trip driving at a steady rate of 25 miles per hour. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She drives at a steady rate of 50 miles per hour. How long after Sheila leaves home will Allison catch up? Distance = rate x time Sheila's rate = 25 mph Sheila's time = t Sheila's distance = 25t {distance = rate x time} Allison's rate = 50mph Allison's time = t  2 {she started 2 hours later or lost two hours off of the starting time} Allison's distance = 50(t  2) = 50t  100 {distance = rate x time} 25t = 50(t  2) {when Allison catches up with Sheila the distances are equal} 25t = 50t  100 {used distributive property} 25t = 100 {subtracted 50t from each side} t = 4 {divided each side by 25} {t corresponds with Sheila's time, which is what the question is asking for} In 4 hours after Sheila leaves Allison will catch up with Sheila  Algebra House
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