Hello all my question is: A student drives 100 mi back tocampus agter spreing br

Question

Hello all my question is: A student drives 100 mi back tocampus agter spreing break and travels with an average speed of57mi/h for 1 hour and 30 minutes. a) what distance was traveled during this time? _______mi b) Traffic gets heavier, and the last part of the trip takesanother half-hour. What was the average speed during this leg ofthe trip? ________mi/h c) Find the average speed for the total trip.________mi/h can someone explain to me the answer and how do i go aboutgetting it for the next time. Thanks Hello all my question is: A student drives 100 mi back tocampus agter spreing break and travels with an average speed of57mi/h for 1 hour and 30 minutes. a) what distance was traveled during this time? _______mi b) Traffic gets heavier, and the last part of the trip takesanother half-hour. What was the average speed during this leg ofthe trip? ________mi/h c) Find the average speed for the total trip.________mi/h can someone explain to me the answer and how do i go aboutgetting it for the next time. Thanks

Explanation / Answer

a) t1 = 1 hour and 30min = 1.5 hour distance d1 = v*t1 = 57 mi/h * 1.5 h = 85.5 miles b) t2 = half hour = 0.5h, distance d2 - 100 miles - 85.5 miles =14.5 miles speed v2 = d2/t2 = 14.5 miles / 0.5 h = 29 mi/h c) total time t = t1+t2 = 1.5+0.5 = 2.0h average speed v = 100 miles / 2.0 h = 50.0 mi/h .

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