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Homework Help: Elementary Mechanics Problem

  1. Sep 12, 2005 #1
    Not sure if this is the correct place to post (not homework because I am studying maths and mechanics in my own time...not a general maths problem either) but...

    I have been working on a problem for a while now that has not only kept me up into the early hours of the morning, but has also woken me up at an early hour, and is just getting frustrating now :frown: The problem is as follows:

    "A body moving in a straight line with a constant acceleration takes 3 seconds and 5 seconds to cover two succsessive distances of 1m. Find the acceleration (Hint: use distances of 1m and 2m from the start of the motion)"

    What I know is the distance covered at two different points of time...I do not know the starting or ending velocity. I have tried numerous approaches (of which most result in garbage) but the one I'm rolling with so far is as follows:

    Though I don't know the velocity(v), I can express it in terms of u
    s=1/2(v+u)t can be transposed to v=(2s/t)-u

    when t =3...v = (2/3)-u
    when t =8...v2 = (4/8)-u

    a = (v-u)/t...or a = (v2-v)/t
    a = ((4/8-u) - (2/3-u))/8-3
    a =(-4/24)/5...=-1/30m/s^2

    problem is that the answer in the book is given as -1/15m/s^2...this answer is twice mine...Where am I going wrong? Is there a blatantly obvious flaw in my understanding that makes my working look ridiculous or just a little something I am missing...please help :confused:
    Last edited: Sep 12, 2005
  2. jcsd
  3. Sep 12, 2005 #2


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    Question: In your original substitution, why do you use [tex]s = \frac{1}{2} (v+u) t[/tex]? Where did the 1/2 come from? Distance is simply [tex]s = vt[/tex].
  4. Sep 12, 2005 #3

    Physics Monkey

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    I think your answer is the correct one, the book appears to be in error. I solved the problem in two different ways and got a = -1/30 m/s^2 both times. Here's one way I did it:

    [tex] \Delta x = v_0 t + \frac{1}{2} a t^2 [/tex]

    This equation yields the following system of equations:

    [tex] 1 = 3 v_0 + \frac{9}{2} a [/tex]

    [tex] 2 = 8 v_0 + \frac{64}{2} a [/tex]

    These can be solved to give the result a = -1/30 m/s^2 (note that I have dropped the units in my equations since everything is in SI). It can be tough to accept sometimes, but the book can be wrong. This is one of those times, unless I made a serious error with both my methods and you're wrong too.


    The equation [tex]s = \frac{1}{2} (v+u) t[/tex] comes from the fact that the motion is accelerated. Put in [tex]v = u + at[/tex] (I'm assuming u is the initial velocity) and you will see that you get [tex]s = ut + \frac{1}{2} a t^2[/tex]. Your equation works if the acceleration is zero (or if you interpret v as the average velocity).
    Last edited: Sep 12, 2005
  5. Sep 12, 2005 #4
    Thanks for taking the time to reply...that your answer is the same as mine with two methods is quite a relief :smile: gonna see if I can answer the rest of the questions now
  6. Sep 12, 2005 #5


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    The confusion is with your interpretation of where the object is at the times stated. The total time to go 2 meters is 5 seconds, not 8 seconds. The question (the way I read it) says that it takes 2 seconds to travel the first meter and then another 3 seconds to go the next meter.

    I too set up 2 equations and two unknowns like Monkey did using [tex]\Dela x = V_oT + \frac{1}{2}aT^2[/tex]:

    [tex]\Delta X_1:[/tex]

    [tex]1 = V_{o1}(3) + \frac{1}{2} a (3^2)[/tex]

    [tex]1 = 3V_{o1} + \frac{9}{2}a[/tex] EQUATION #1

    [tex]\Delta X_2:[/tex]

    [tex]2 = V_{o2}(5) + \frac{1}{2} a (5^2)[/tex]

    [tex]2 = 5V_{o2} + \frac{25}{2}a[/tex]

    [tex]-5V_{o2} = \frac{25}{2} a -2[/tex]

    [tex]V_{o2} = \frac{-5}{2} a + \frac{2}{5}[/tex] EQUATION #2

    Now plug in Equation 1 into EQUATION 2:

    [tex]1 = 3(\frac{-5}{2} a + \frac{2}{5}) + \frac{9}{2} a[/tex]

    [tex]1 = \frac{-15}{2} a + \frac{6}{5} + \frac{9}{2} a[/tex]

    [tex]\frac{-1}{5} = -3 a[/tex]

    [tex]a = \frac{1}{15}[/tex]

    I must have messed up a sign in there somewhere, because it should end up negative because it is slowing down. Let me see if I can find the goof...
    Last edited: Sep 12, 2005
  7. Sep 13, 2005 #6
    Thanks for the reply Fred Garvin...I can now see where the answer of -1/15m/s^2 is coming from...when you substituted 'v' for 'u' in the equation s=1/2(v+u)t I think you should end up with s = vt-(1/2)at^2..because u=v-at


    1=3((5/2)a +2/5)-(9/2)a
    1=(15/2)a +6/5 -(9/2)a
    -1/5 =3a
    a=-1/15 :biggrin:

    thanks a lot folks...my natural impression is that... "3 seconds and 5 seconds" out of 5 seconds reads as 3 seconds then a further 2 though!...which would be acceleration (hmph...naughty textbook tries tricks us)
    Last edited: Sep 13, 2005
  8. Sep 13, 2005 #7

    Physics Monkey

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    Ok, I'm confused here. Fred, you seem to be interpreting the problem as saying that it takes 3 seconds to go the first 1 m and then 5-3=2 seconds to go the second meter (or 5 seconds the go the whole 2 m). This is different from what you say at the top of your post. I base this on the fact that you use a time of 3 (rather than 2) in your first equation and you equate [tex] v_{0 1}[/tex] and [tex] v_{0 2} [/tex] so you're starting from the same point in both cases. You didn't make a mistake in your algebra after that point. Indeed if it takes 3 seconds to cross the first 1 m and only 2 seconds to cross the second 1 m then the acceleration must be positive, it must be speeding up.

    I really think the problem is saying that it takes 3 seconds to go the first 1 m and then 5 seconds to go the second 1 m (for a total of 8 seconds from the start). If the problem doesn't mean this then I think it is very poorly worded. Thoughts?
  9. Sep 13, 2005 #8


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    I agree that the problem is poorly worded. I went with it because that was the way it made sense to me at first. The more times I read the problem statement, the more confusing it seems. I have a sneaky suspicion that the authors had a similar thing happen to them.
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