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Average Speed Over Given Distance - Driving me Nuts

  1. Feb 11, 2015 #1
    "A car accelerates from rest over 400 metres in 19 seconds. The driver then brakes and the car stops in 5.1 seconds with constant deceleration."

    Part one of the question is "calculate the acceleration for the first 400m" which went just fine:

    x=ut+(1/2)at^2
    400=180.5a
    a=2.22 m/s

    Now the second part of the question asks for the average speed over the entire journey, both accelerating and braking sections. So this is what I did:

    Final velocity after 400m:

    x=(1/2)(u+v)t
    400=(1/2)(0+v)19
    v=42

    It's not exactly 42 but I did use the exact answer for the rest of this. Anyway, I then did:

    Braking distance:
    x=(1/2)(42+0)5.1
    x=107.1

    So it should be 507.1/24.1=21 m/s

    But when you do an acceleration/time graph and solve the area under the line you get the apparently correct answer of 26 m/s. I cannot for the life of me understand why or how my method doesn't work.
     
  2. jcsd
  3. Feb 11, 2015 #2

    BvU

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    The acceleration vs time graph ? That area gives you the speed at 24.1s which is 0.
    Can you show how you solve the area under the line ?

    I think you're doing fine with your calculations and should question the 'apparently correct result' !

    Oh, and: hello Sarlacc, Welcome to PF :)
     
  4. Feb 11, 2015 #3
    A friend of mine did it with a graph. at 19s the acceleration is 2.22. then it goes down to zero which takes 5.1 seconds.
    half-base*height for the first side:

    (1/2)(19)*2.22= 21
    and then
    (1/2)(5.1)*2.22=5.66

    Which is of course 26-ish in total.

    That works. And my teacher said it's the right answer, he tried my method and after seeing my friend's work he said I'd calculated the "distance". So that wasn't much help at all.

    I don't doubt that it's the right answer. I just don't get why the method I used doesn't work. And I've checked the values with every equation I can think of.
     
  5. Feb 11, 2015 #4

    BvU

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    The acceleration is a constant (we assumed that) 2.22 for 19 seconds. Then it is a constant -8.26 for 5.1 seconds. Area 42.105 - 42.105 = 0. No triangles involved.

    Now the speed graph: it goes from 0 to 42.105 (slope 19 * 2.21) in 19 seconds. Triangle area: base * half height = 400 meter (we knew that already). Then it goes down from 42.105 to 0 in 5.1 seconds. Triangle area 107.4 m (as you already calculated correctly). Total 504.7 m which (of course!) is a distance.

    And sure enough, average speed is area/width, so 507.4/24.1

    If "A car accelerates from rest over 400 metres in 19 seconds. The driver then brakes and the car stops in 5.1 seconds with constant deceleration" is correct, then so are you.

    Teachers are very busy people, so sometimes perhaps they react a bit too quickly and draw the wrong conclusions ?
     
    Last edited: Feb 11, 2015
  6. Feb 11, 2015 #5
    dodgygraph.jpg

    That's two triangles?
     
  7. Feb 11, 2015 #6

    BvU

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    Yes, that's two triangles. But it's a nonsense graph. It certainly doesn't represent a ! a doesn't start at zero and then gradually increases. And it it isn't +2.22 when the braking has started and then gradually goes to zero again. See post #4. The graph of a is trivial. Two rectangles, one above, one below the t axis.

    And to boot: there's no way you can get average v from an a graph (at least not directly). Average a, yes.
    And that average a is zero: v was zero at start, is zero at end.

    Oh, and: bring it tactfully. Teachers (are humans) need some time to adjust when wrong...
     
  8. Feb 11, 2015 #7
    Wha?

    The question states that the car accelerates from rest (0) over 400m for 19 seconds. And then it decelerates at a constant rate for 5.1 seconds.

    OHHHHHHHHHHH I think I see what you're saying. I'll try asking him that tomorrow.
     
    Last edited: Feb 11, 2015
  9. Feb 11, 2015 #8

    haruspex

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    Your friend's method assumes, with no justification, that the deceleration is the same magnitude (2.22 m/s2) as the acceleration. Correspondingly, he/she has neither used nor checked that the vehicle comes to rest in the 5.1 seconds.
    Worse, your friend then adds together the two calculated average speeds. That is nonsense.
    Your method is entirely correct and gives the correct answer. In fact, you can get there a little more quickly. Having found the peak speed to be 42 m/s, you can argue that since it is constant acceleration between that speed and rest in each phase, the average speed is 21 m/s in each phase, so the overall average speed is 21 m/s.
     
  10. Feb 11, 2015 #9
    I have class tomorrow. Hopefully I'll find out the final, definite, for sure, no-really answer then.
     
  11. Feb 11, 2015 #10
    Turns out I was right! I knew it'd come down to me not noticing something simple, so thanks for pointing out how that a/t graph didn't make sense.

    Thanks for the help guys!

    I'd edit the title and stick this update in my last point but looks like I can't. But I just wanted to let anyone interested know it's case closed.
     
  12. Feb 12, 2015 #11

    BvU

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    Good learning experience and confidence builder !
     
  13. Feb 12, 2015 #12

    PeroK

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    Here's something interesting. The answer you got was 21 m/s, which just happens to be 400m/19s. Is that a coincidence?
     
  14. Feb 12, 2015 #13
    Huh. That's pretty cool. I think it's just a coincidence though. Maybe.
     
  15. Feb 12, 2015 #14

    PeroK

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    Maybe not! What can you say about the average speed when accelerating uniformly from rest, or decelerating uniformly to rest?
     
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