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32 votes
32 votes
A student drives the 100-mi trip back to campus after spring break and travels with an average speed of 65 mi/h for 1 hour and 30 minutes for the first part of the trip.(a) What distance was traveled during this time? mi(b) Traffic gets heavier, and the last part of the trip takes another half-hour. What was the average speed during this leg of the trip? mi/h(c) Find the average speed for the total trip. mi/h

User Brosto
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1 Answer

12 votes
12 votes

a) The formula for calculating the distance is expressed as

distance = speed x time

From the information given,

speed = 65 mi/h

time = 1 hour and 30 minutes = 1.5 hors

Distance = 65 x 1.5

Distance during this time = 97.5 miles

b) Recall,

Total distance = 100 miles

Distance covered in the last part of the trip = 100 - 97.5 = 2.5

Time = half-hour = 0.5

Speed = 2.5/0.5 = 5

the average speed during this leg of the trip = 5 mi/h

c) average speed = total distance/ total time

total time = 1.5 + 0.5 = 2 hours

total distance = 100

average speed = 100/2 50

Average speed for the total trip is 50 mi/h

User JDC
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