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Uniformly Accelerated Linear Motion

  1. Feb 9, 2008 #1
    [SOLVED] Uniformly Accelerated Linear Motion

    Hello,

    I am attempting what looks to be a fairly easy problem involving uniform acceleration and linear motion, but my answers just aren't coming out right and I'm unable to see where I'm going wrong. Any help is appreciated.

    1. The problem statement, all variables and given/known data

    A teenager has a car that accelerates at 3.00 m/s[tex]^{2}[/tex] and decelerates at -4.50 m/s[tex]^{2}[/tex]. On a trip to the store, he accelerates from rest to 15.0 m/s, drives at a constant speed for 6.00 s, and then comes to a momentary stop at the corner. He then accelerates to 18.0 m/s, drives at a constant speed for 15.0 s, and decelerates for 2.67 s. Then he continues for 4.00 s at this speed and comes to a stop.

    (a) How long does the trip take?
    (b) How far has he traveled?


    2. Relevant equations

    (i) [tex]\Delta[/tex]X = [tex]\frac{v^{2}_{f} - v^{2}_{i}}{2a}[/tex]

    (ii) [tex]\Delta[/tex]X = [tex]v_{i}t[/tex] + [tex]\frac{1}{2}at^{2}[/tex]

    (iii) [tex]v = v_{i} + at[/tex]


    3. The attempt at a solution

    I've broken the problem down into the 6 parts of the trip and calculated both the time and distance over each and then added them up for the total.

    (a) Calculating time taken

    Part I: Accelerates from rest to 15.0 m/s, so:
    [tex]v_{i}[/tex] = 0 m/s, [tex]v_{f}[/tex] = 15.0 m/s, a = 3.00 m/s[tex]^{2}[/tex]​

    Solving equation (iii) for t and plugging in values gives:
    t = (15.0 - 0)/3.00 = 5.00 s.​

    Part II: Constant speed for 6.00 s, so no calculation needed; t = 6.00 s.

    Part III: After momentary stop, accelerates from rest to 18.0 m/s, so same as part I.
    [tex]v_{i}[/tex] = 0 m/s, [tex]v_{f}[/tex] = 18.0 m/s, a = 3.00 m/s[tex]^{2}[/tex]​
    t = (18.0 - 0)/3.00 = 6.00 s.​

    Part IV: Constant speed for 15.0 s, so again no calculation needed; t = 15.0 s.

    Part V: Decelerates for 2.67 s; time given, no calculation needed; t = 2.67 s.

    Part VI: Constant speed for 4.00 s, no calculation needed; t = 4.00 s.

    Adding up time taken for all parts: 5.00 s + 6.00 s + 6.00 s + 15.0 s + 2.67 s + 4.00 s = 38.67 s.

    (b) Calculating distance traveled

    Part I: Using equation (i) with [tex]x_{i}[/tex] = 0, [tex]v_{f}[/tex] = 15.0 m/s, [tex]v_{i}[/tex] = 0 m/s, a = 3.00 m/s[tex]_{2}[/tex],
    X = (15.0)[tex]^{2}[/tex]/2(3.00) = 37.5 m​

    Part II: Constant speed of 15.0 m/s for 6.00 s, so X = 15.0 * 6.00 = 90.0 m.

    Part III: Same as part I after stop, so using equation (i) with [tex]x_{i}[/tex] = 0, [tex]v_{f}[/tex] = 18.0 m/s, [tex]v_{i}[/tex] = 0 m/s, a = 3.00 m/s[tex]^{2}[/tex],
    X = (18.0)[tex]^{2}[/tex]/2(3.00) = 54.0 m​

    Part IV: Constant speed of 18.0 m/s for 15.0 s, so X = 18.0 * 15.0 = 270 m.

    Part V: Decelerates for 2.67 s, so using equation (ii) with [tex]v_{i}[/tex] = 18.0 m/s, t = 2.67 s, a = -4.50 m/s[tex]^{2}[/tex],
    X = (18.0)(2.67) + (0.5)(-4.50)(2.67)[tex]^{2}[/tex] = 32.0 m​

    Part VI: Constant speed for 4.00 s, so first need to find [tex]v_{f}[/tex] after deceleration in part V by using equation (iii) with [tex]v^{i}[/tex] = 18.0 m/s, a = -4.50 m/s[tex]^{2}[/tex], and t = 2.67 s.
    v = (18.0) + (-4.50)(2.67) = 5.99 m/s​

    Travels at constant speed of 5.99 m/s for 4.00 s, so 5.99 * 4.00 = 23.9 m.

    Adding up all of the distances gives total: 37.5 m + 90.0 m + 54.0 m + 270 m + 32.0 m + 23.9 m = 507.4 m.

    I must be missing something that's glaringly obvious here. I thought perhaps the undefined "momentary stop" might be skewing my results for part (a), but more likely I've overlooked something. Thanks in advance to whoever can point out where I've gone wrong.
     
  2. jcsd
  3. Feb 9, 2008 #2
    I wonder how he managed to come to that momentary stop, without decelerating ?
     
  4. Feb 9, 2008 #3
    >>Then he continues for 4.00 s at this speed and comes to a stop.

    I think there will be a Part VII for calculating the distance traveled while decelerating.

    x=(0^2-5.99^2)/(2*(-4.50))


    How do you type the square ? I don't know how to get that square so i use ^
     
  5. Feb 9, 2008 #4
    Thanks. I knew I was excluding something. I just factored in those 2 stops and got the answer.

    I used the Latex code this forum has. There is a reference for all of the codes on the post message page. You can click on the sigma icon and then "Subscript and Superscript," then the picture showing a superscript and type a '2' between the brackets. The code is something like x[ tex]^{2}[ /tex] without the spaces to get x[tex]^{2}[/tex]. A lot faster to just type ^ though.
     
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