SOLUTION: Please help me solve this question: A student drives the 100 mi trip back to campus after spring break and travels with an average speed of 54 mi/h for 1 hour and 30 minutes. There

Question 569224: Please help me solve this question: A student drives the 100 mi trip back to campus after spring break and travels with an average speed of 54 mi/h for 1 hour and 30 minutes. There are three parts the first part asks for the distance traveled during this time in miles. The second one says that the traffic get heavier on the last part of the trip so it takes another half-hour and it wants to know the average speed of this leg of the trip (mi/h. The final part asks for the final average speed for the total trip (mi/h).
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A student drives the 100 mi trip back to campus after spring break and travels with an average speed of 54 mi/h for 1 hour and 30 minutes.
There are three parts
:
the first part asks for the distance traveled during this time in miles.
Dist = speed * time
d = 54(1.5)
d = 81 mi
;
The second one says that the traffic get heavier on the last part of the trip
so it takes another half-hour and it wants to know the average speed of this leg of the trip (mi/h.
Speed = dist/time
time = .5 hr; dist: 100 - 81 = 19 mi
S =
S = 38 mph speed on the last part of the trip
:
The final part asks for the final average speed for the total trip (mi/h).
Let a = average speed, total time: 1.5 + .5 = 2hrs
Write a time equation: time = dist/speed
1.5 + .5 =
2.0 = 100/a
2a = 100
a = 100/2
a = 50 mph was the average speed for the whole trip
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