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Kinematics Problems

  1. Aug 21, 2007 #1
    I just wanted to double check on my answers here. I have all the problems posted below with my work. If you could make sure I'm doing these right I would greatly appreciate it.

    1. The problem statement, all variables and given/known data
    A truck passes a point A on a straight level road with a constant velocity of 10 m s-1. At the same instant, an automobile starts from rest at point A and travels in the same direction as the truck, with a constant acceleration of 1.6 m s-2.

    A) How long a time does the automobile take to pass the truck, from the instant it starts at point A?

    B) How far beyond the starting point will the automobile overtake the truck?

    C) Determine the average speed of the bus during the interval, t=0 to t=10 s.

    D) How fast will the automobile be traveling, when it passes the truck?

    E) If the speed limit on this road is 50 km/hr., is the automobile within the speed limit as it passes the truck?


    My work is attached below. I did a problem similiar to this and found that the answer for my time was half what it should have been.This messed up all my other answers. I'm not sure where I'm making my mistake in finding the time.
     

    Attached Files:

  2. jcsd
  3. Aug 21, 2007 #2
    I can't see your attachment yet (pending approval), but the time for question A) should be 12.5s. What did you get?
     
  4. Aug 21, 2007 #3
    I got 12.5 once i doubled my answer. I'm not sure why. I used the formula final velocity = oringal velocity + accel. x time

    that got me 6.25 (doubled is 12.5 sec)

    my other answer were

    B) 125m
    C) 16 m/s
    D) 20 m/s
    E) yes
     
  5. Aug 21, 2007 #4
    The equation to find the time should be

    [tex]\frac{1}{2}at^2 = vt[/tex]

    Leading to the solution t = {0,2v/a}
    You probably forgot the 1/2 in the equation for accelerated motion.

    Looks like you are also forgetting that factor of 1/2 for question C.

    You might want to try question E again. What is 50km/h in m/s?
     
  6. Aug 21, 2007 #5
    For C I should find the distance first whic is 80m in the first10 sec

    then use v = d/t

    then that gives me 8m/s

    That would mean for E

    50 km/h = 180 m/s
    so that would make the answer No


    The answers look good though?
     
  7. Aug 21, 2007 #6
    8 m/s is correct. Everything else looks good except E. 50 km/h is not 180 m/s.

    Use dimensional analysis

    (50km/h)(X m/km)(h / X min)(min / X s)

    Fill in the X's, and multiply through to get your answer in m/s.
     
  8. Aug 21, 2007 #7
    13.8 m/s
    So that would be yes
    I think I'm better at converting now that you broke it down for me.

    Do you think you could look at one more problem. I think I might have did a little better on it but I just want to be sure. If you dont have time o problem. Thanks

    (I'll just post my scanned work)
     
  9. Aug 21, 2007 #8
    Sure thing.
     
  10. Aug 21, 2007 #9
    A ball is thrown vertically upward from the top edge of a building 250 m tall with an initial velocity of 25 m/s. Assume the acceleration due to gravity is g = 10 m s-2.

    A) How long will it take for the ball to rise to the highest point on its trajectory?

    2.5 sec

    B) What is the maximum height, measured from the ground, to which the ball rises?

    281.25m

    C) On its way up, when is the speed of the ball equal to half its speed of projection ?

    1.25 sec

    D) Assuming that the ball misses the building on its way down, at what time does the ball hit the ground ?

    5 sec

    E) What is the magnitude of the velocity with which the ball hits the ground?

    55 m/s (I got this answer from D, I couldn't fit it on the page. This comes from V2)
     

    Attached Files:

  11. Aug 21, 2007 #10

    HallsofIvy

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    You have, for "A)How long a time does the automobile take to pass the truck, from the instant it starts at point A?",
    [itex]v_f= v_0+ at[/itex]
    10 m/s= 1.5 m/s2 t
    t= 6.25
    and then say "I used 12.5 because that's 2 times 6.25"! Without giving any reason for doubling! Even if it is true that you got 1/2 the correct value before, that is no reason for arbitrarily doubling whatever you get!

    In any case, you are answering the wrong question! vf= v0+ at says two speeds are the same. The answer you got is the time after which the two vehicles have the same speed, not the same distance, which is what is asked. Since the car starts out going slower than the truck, at the time the speeds are finally the same, the car will be well behind the truck still.

    Yes, you could argue, since this is a constant speed versus a constant acceleration, the distance the car loses while it is slower than the truck will be offset by the distance the car gains while it is faster than the truck- by "symmetry" the total time should be twice the time until they have the same speed- but at least you know now why you multiply by 2 (and when that will not work).

    A better, clearer way to do this is to use distance formulas rather than velocity formulas. The truck, moving at constant vt will go vt t meters in t seconds. The car, at constant acceleration, will go a distance of (1/2)at2 meters in t seconds. The car and truck will have gone the same distance when vt t= (1/2)at2. Notice that that has the obvious solution t= 0. If t is not 0, we can divide both sides by t to get vt= (1/2)at. Now put in the numbers you are given and you will get the correct answer- but do you see why you had to double your first answer?

    "B) How far beyond the starting point will the automobile overtake the truck?"
    Once you have the answer from (A), put that value of t into either side of the equation for distance. I notice you used (1/2)at2. You would get the same answer, and the arithmetic is easier, using vt t.

    "C) Determine the average speed of the bus during the interval, t=0 to t=10 s."
    I have no idea how to answer this since there is no bus in the problem! Assuming you meant "automobile", divide the distance it went that you got in (B) by the time you got in (C). For some reason, you use 10 seconds as the time, even though you got 25 seconds as the time in (A). What you have calculated is the speed of the automobile after 10 seconds. I have no idea why you think that would be the average speed.

    "D) How fast will the automobile be traveling, when it passes the truck? "
    Put the time you got in (A) into the formula for speed with constant acceleration. What you have is correct.

    "E) If the speed limit on this road is 50 km/hr., is the automobile within the speed limit as it passes the truck?"
    You multiplied 20 m/s by 3.6 and got 72 km/hr and answered "Yes".
    First you should say WHERE you got that "3.6". It is (almost) correct but you should show where you got it.
    The reason I said "almost" is that "20 m/s times 3.6" is NOT "72 km/hr"- it is 72 m/s. 20 m/s times 3.6 (km/m)(second/hr) is 72 km/hr.

    Finally, if you think going 72 km/hr is "within the speed limit" of 50 km/hr then I don't want to be on the same road with you!
     
  12. Aug 21, 2007 #11
    Everything looks good except for E.

    The equation for V(t) is V(t) = 25m/s - (10m/s)t
    So evaluate the magnitude of V(5)
     
  13. Aug 21, 2007 #12
    So the magnitude for E would be 3?
     
  14. Aug 21, 2007 #13
    Sorry, it looks like your answer for D is also wrong. Which of course screws up the answer for E. I will wait until I can see your work.
     
  15. Aug 21, 2007 #14
    Thanks for the explanation halls. I just needed someone to explain exactly that for me. Now I'm coming up with 15 s for A.

    I did:

    12 m/s (t) = .5 (1.6 m/s) t^2
    t = 15 sec
     
  16. Aug 21, 2007 #15
    Why 12m/s?

    "A truck passes a point A on a straight level road with a constant velocity of 10 m s-1"
     
  17. Aug 21, 2007 #16
    ok, I got it mixed up with another problem
     
  18. Aug 21, 2007 #17
    for problem 2 here's my work

    final V = original V + a x delta T

    V2= 30 m/s + 10 m/s^2 x 2.5s
    V2 = 55 m/s

    then i used the formula:

    t = V2-V1 / a

    t = (55 m/s - 30 m/s) / 10 m/s^2
    t = 2.5 sec

    I multiplied that by 2 since if took 2.5 to reach the top
     
  19. Aug 21, 2007 #18
    After running through problem 1 again I got:

    A) 12.5 sec
    B) 125 m
    C) 12.5 m/s
    D) 20 m/s
    E) NO

    Is that more like it. Thanks for the help.
     
  20. Aug 21, 2007 #19
    Ok there's your problem. You are using a to be +10m/s^2. However, you throw the objects up, and gravity pulls it down, so you want a negative sign. A more suitable equation of motion would be

    y(t) = 250 + 25t - (1/2)(10)t^2
    v(t) = 25 - 10t

    Using this for D, you can use y(t) = 0 (when it hits the ground), then find the time it takes. For E, substitute that value into the v(t) equation.
     
  21. Aug 21, 2007 #20
    I think I found my mistake.

    t = (281.25 m/s) / (10m/s^2)
    t = 28.125 sec

    28.125 sec + 2.5 sec = 30.625 sec

    That would make E:

    V2 = V1 + a(delta t)
    v2= 10 m/s^2 (28.125 sec)
    v2 = 281.25 m/s
     
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