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Can some please help me on this problem?? I think I’m setting it up wrong...

Can some please help me on this problem?? I think I’m setting it up wrong...-example-1

1 Answer

4 votes

Answer:

60 miles per hour

Explanation:

Let s represent the speed in miles per hour during the first part of the trip. Then s-10 will be the speed during the last part of the trip.

For speed/time/distance problems, the appropriate relation is the one that is posted on every speed limit sign:

speed = miles/hour = distance/time

Rearranging this relation gives you the expression for time:

time = distance / speed

You are given time and distance and you need to find speed. The time you're given is the total for the two parts of the trip, so ...

4 = 84/s + 130/(s-10) . . . . . . total time = time1 + time2

Multiplying by the product of the denominators, this becomes ...

4s(s-10) = 84(s-10) +130s

4s^2 -254s +840 = 0 . . . . . . subtract the right side to put in standard form

This can be solved using any of several methods for solving quadratic equations. Solutions are ...

s = 60, s = 3.5 . . . . . only the first solution makes any sense in the problem

The speed during the first part of the trip is 60 miles per hour.

User Lee Warner
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