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Having Trouble with these 2 problems

  • Thread starter buck3y3nut
  • Start date
  • #1
Having Trouble with 1 problem now. Plz help! Thank you!

Hey guys,
This might be cake for you, but it's bugging me right now because I can't figure it out...

I figured out the 1st problem. Trying my hardest to get this one. Can someone please give me a hint on how to approach this problem please?

2. Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 20 mph and half the distance at 60 mph. On her return trip, she drives half the time at 20 mph and half the time at 60 mph.
(a) What is Julie's average speed on the way to Grandmother's house?

I got the right answer. Just had to convert the .5 to minutes and i came up with 30 mph for this answer which was right... Now I can't figure out the 2nd part...

(b) What is her average speed on the return trip?

Any hints on how to approach this problem?

Any help will be greatly appreciated. Sorry, never took physics in high school so this might be cake for some of you...
Thank you again
 
Last edited:

Answers and Replies

  • #2
Can someone please help me???
 
  • #3
HallsofIvy
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You say you found it easy to solve the first problem? I'm wondering what you mean by "convert the .5 to minutes"?? "Minutes" don't come into at all!

Yes, 30 mph is the correct answer. Here is how I did it: She drove half the trip, 50 miles, at 20 mph- that would take her 50/20= 5/2 hours. She drove half the trip, 50 miles, at 60 mph- that would take her 50/60= 5/6 hour. The entire trip, 100 miles, took her 5/6+ 5/2= 5/6+ 15/6= 20/6 hours. Going 100 miles in 20/6 hours means her average speed was 100/(20/6)= 6(100/20)= 6(5)= 30 mph.

On the way back 1/2 of the time she drove 20 mph and 1/2 of the time she drove 60 mph. Okay, that is a little harder because we don't know the total time she took and, unlike the first problem, we don't know the two distances. We can, however, say call the distance she went at 20 mph d1, the distance at 60 mph d2, and the total time she took T. Then we can say, since she drove distance d1 at 20 mph in T/2 hours, d1= 20(T/2)= 10T and, since she drove distance d2 at 60 mph in T/2 hours, d2= 60(T/2)= 30T.
Now, since we know the total distance was 100 miles, add those two equations:
d1+ d2= 100= 10T+ 30T= 40T. T= 100/40= 5/2 or 2 and a half hours. Her average speed was 100/(5/2)= 2(100/5)= 40 mph.\

Notice, by the way, that in the second problem, specifically because it is time that is equal, her average speed is just the arithmetic average of the two speeds.
 
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