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Tricky velocity/time problem without distance

  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22-min rest stop. If the person's average speed is 77.8 km/h
    a) How much time is spent on the trip?
    b) How far does the person travel?

    2. Relevant equations
    v1 = 89.5 km/h -> t
    v2 = 0 km/h -> 0.37 h
    AvgV = 77.8 km/h
    a = 0 since speed is constant

    3. The attempt at a solution
    I have been given formulas throughout my textbook to solve for displacement, velocity, acceleration, and things like this. I will try to briefly show the kind of problems I run into and my thought process in solving this:

    I'm looking for x (distance) and t (time) anywhere I can make it work basically.

    average v = Δx / Δt I don't have Δx so I can't use this to find t.

    v = vo + at I have original velocity but I can't isolate time because a=0.

    All of my other equations don't work because they divide by 0, there are 2 unknown variables, or other such problems.

    Im starting to wonder about how my "t"s are split into 2 parts in my question (0.37h and unknown) but in my equations there is only one variable t. I don't think my equations will work.

    I can average the two velocities out to 89.5 km/h, but I can't figure out how to account for times which would make one velocity count for more than the other.

    I tried to input 1h for my unknown t and calculate average velocity just to see if that would help me better understand the problem. I still don't know how to calculate average velocity without x (displacement). And I don't know how to find x with two different time values. Im really stuck!
     
  2. jcsd
  3. Oct 12, 2015 #2

    billy_joule

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    The trip is made up of two legs, driving (1) and resting (2), combining the information from both legs is what you need to do to solve:

    xtotal = x1 +x2

    ttotal = t1 + t2 = t1 +0.37 h

    express x1 in terms of t1 and V1 and then you can solve:

    Vaverage = xtotal / ttotal
     
  4. Oct 12, 2015 #3

    collinsmark

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    Hello @PhysicsBoyMan,

    Welcome to Physics Forums! :smile:

    You'll use that formula, so keep it handy. You'll actually use it twice. But before you do, I suggest defining your terms a little better (see below).

    Yeah, don't worry about acceleration. It doesn't matter for this problem.

    Yes, you do need to split something apart. So as part of that split, start defining terms specifically. I suggest something along the lines of the following:

    [itex] v: [/itex] velocity while driving.
    [itex] v_a: [/itex] average velocity.
    [itex] \Delta t: [/itex] driving time.
    [itex] \Delta t_r: [/itex] resting time.
    [itex] \Delta t_t: [/itex] total time.
    [itex] \Delta x: [/itex] distance

    Now see if you can simplify, and solve for the driving time, [itex] \Delta t [/itex], as an interim step. (Hint: You'll use your formula a couple of times in the process.)
     
  5. Oct 12, 2015 #4
    Thanks for the help. I expressed x1 in terms of t and v like you said. I have two occurences of t1. Is that normal? Is this ok so far? I tried to continue this path a few different ways but failed to come up with a single t value.

    photo.jpg
    upload a picture
     
  6. Oct 12, 2015 #5

    billy_joule

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    What you have so far is correct.
    If you multiply both sides by (t1 + 0.37 h) you'll get:

    (77.8km/hr) (t1 + 0.37 h) = (89.5km/hr)(t1)

    expand the LHS then solve for t1
     
  7. Oct 13, 2015 #6
    I don't know how to isolate one occurrence of t1 when there is multiplication involved. Here are a few methods I tried and failed with. I can't imagine any other possible way.
    photo.jpg
    click image upload

    Can someone please help me finish this it would be useful to learn how to isolate a variable when there are two of the same.
     
  8. Oct 14, 2015 #7

    haruspex

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    I don't understand your second line. What happened to the +0.37h?
    Starting again from the first line, get all terms with a t1 factor on one side of the equation and all terms without it on the other.
     
  9. Oct 14, 2015 #8
    What happened to the +0.37h in the second line is that Billy Joule said to expand the LHS and solve for t1, so I multiplied (77.8km/h) (t1+0.37h).

    So back to the first line. I tried to get all terms with t1 factor on one side. This attempt is not shown in my picture, but I found that dividing both sides by (t1+0.37h) left me with a t1 divided by a t1 which equals 1 and I lose my t1. So besides dividing, how else can I get both occurrences of t1 to the same side?

    In my second method in my picture (bottom blue box), I tried to take the whole right side (89.5km/h)(t1) and subtract that from both sides. That's why it is in square brackets and negative on the LHS. Then I tried to factor out t1 from that.

    Then in lines 2 and 3 in the first blue box (I know my work is scattered all over the map) I tried to simply first to see if I would be able to gather t1 terms on the same side later. This did not work.

    I tried to divide, I tried to subtract, but I was unable to get both t1 terms on the same side of the equation in the same numerator.
     
  10. Oct 14, 2015 #9
    I just had a bit of a breakthrough. If I can divide (km/h)(t1) / (km/h)(t1) then I would be left with a single t1 term. Only problem is that my t1 on the LHS is tied up in an addition equation. I tried to expand it by multiplying but someone said I did it the wrong way.
     
  11. Oct 14, 2015 #10
    [itex](77.8km/h)(t1 + 0.37H) ≠ (77.8km/h)(t1)(28.8km/h) [/itex]

    Rather: [itex] x(y + z) = xy + xz[/itex]
     
  12. Oct 14, 2015 #11
    So here is what I did. My answer sheet says that it should be 2.8h. This is the closest I got. I hope the discrepancy is just from leaving out a few decimal places here and there. Is this correct?

    image.jpg
    photo hosting
     
  13. Oct 14, 2015 #12

    collinsmark

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    Your solution looks about right to me. There is some error due to rounding (most notably involving the resting time), but that's not enough to explain the entire discrepancy. I think there might be a mistake in the answer sheet.
     
    Last edited: Oct 14, 2015
  14. Oct 14, 2015 #13

    haruspex

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    The question wording implies the 22 minutes rest is to be counted as part of the trip.
     
  15. Oct 14, 2015 #14

    collinsmark

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    Right. I'm just saying that 22/60 = 0.37 contains the bulk of the rounding errors. (Not enough to explain the discrepancy with the answer sheet though.)

    [Edit: Oh, wait. I get it. Add in the 22/60 hours to the traveling time to get the total trip time. That comes out to around 2.8 hours, yes. Nevermind.]
     
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