1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Graph and Questions

  1. Feb 16, 2014 #1
    The graph in the picture attached describes the motion of a 500.0 g cart for the first 60 seconds of its motion.
    1. Calculate how far the cart moves during the fi rst 25 seconds of motion.
    2. Indicate the time intervals during which the cart has a non-zero acceleration.
    3. Calculate the acceleration of the cart at 35 seconds.
    4. Indicate all times during which the cart is at rest.
    5. Calculate how far the cart is from its starting location after 60 seconds.
    6. Calculate the additional time required for the cart to arrive back at its initial position if it continues to move at 50 m/s as indicated on the graph.

    My attempt:
    1. area of triangle + area of rectangle = total displacement in first 25 seconds = distance
    1000 m + 500 m = 1500 m
    2. (0,20) and (30,45)
    3. 50-100/5 = -50/5 = -5 m/s^2
    4. At t = 0 and t = 40
    5. Total area covered? 1000 m + 1000 m + 500 m + 125 m + 750 m = 3375 m
    6. I did not understand this question. Do i just get the area (distance) and speed is given. So i can get the time taken? I don't know why they would give the mass of the object?
    I haven't used it answering any of the questions--that kinda puts me in doubt as to if I answered them correctly.

    Can somebody please review them and let me know where I'm wrong?

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Feb 16, 2014 #2
    You have the right idea for all of these I think. 1 - 4 are definitely right.

    For 5, it will be the total area covered, but you have to make the areas below the t-axis negative, since those are places where the cart moves backwards, which would move it closer to the starting location. If it was asking for the total distance traveled you would make each area positive, but for displacement you include signs on all the areas. Those questions can be difficult to answer.

    For 6, I think you need the answer from 5. For 5 you get a positive area representing how far the cart is from the starting point, so 6 is essentially asking how much "negative" area you would need to cancel out the area in 5 completely, if you have the cart moving at -50 m/s as in the graph. You could also use the displacement numerically, and then do a simple calculation with constant to velocity to find how much time is needed to cover the displacement, in the opposite direction of course.

    I don't see a use for mass in any part of the problem, so don't worry about it. Sometimes they'll throw that in there just to trick you, or for the sake of completeness in the problem description.
     
  4. Feb 16, 2014 #3
    @ jackarms You always answer my questions :) Thank you bro.
    Anyways, getting back to the question. I calculated the total distance traveled by the cart. So displacement is direction sensitive, so to speak. Would it help if I converted it into a position vs time graph? I don't think it is needed but I'm still a wee bit confused on the negative displacement part (below the x axis) It does kinda make sense so here's my attempt at 5:
    2500 m - 125 m -750 m = 1625 m
    In other words, the cart is 1625m from it's starting position?
    And now we can use this to find 6:
    D= SxT ; T= D/S
    1625 m/ -50 m/s = 32.5 secs

    Does that sound right?
     
  5. Feb 16, 2014 #4
    Also, I think my original answer to 2. is wrong. Well there's nothing wrong with (0,20), however, (30,45) should be broken down into two parts because the velocity is zero at t=40, which means the cart was at rest for a second?Should I break it into (30,39) and (41,45)? or simply (30,40) and (40,45)?

    This is so confusing.
     
  6. Feb 16, 2014 #5
    And did I mess up + - for 3?
    Could it be -50m-100m/5 secs = -150 m/5 secs= -30 m/s^2?
    My reason is since the curve is downwards the velocity must be decreasing--so it's -50 m/s at t=35 instead of 50 m/s
     
  7. Feb 16, 2014 #6
    Hey, no problem for answering your questions. It's always nice to brush up on my physics too.

    Your calculations of displacement and the time for #6 both look right. And you're right about displacement being direction sensitive. That's the difference between it and distance traveled, and they're certainly easy to mix up. Making a position-time graph is another possibility, but it's difficult since the curve would be parabolic, and to determine all the points you would have to do the area calculations anyway. If you can understand the concepts of positive and negative area for a velocity-time graph, I think that would be easier.

    For #2, your original answer is correct. It's a little confusing how the cart can still be accelerating when it's at rest, but it still is, so you would want to include t = 40 in the answer. You can think of it as the nonzero acceleration changing the velocity from positive to negative just at that point. Another example is if you toss an object up into the air, what will be the acceleration at it's maximum height. It's tempting to say zero since it's at rest, but it's still in free-fall, so it must still be accelerating.

    And for 3, your original answer is also right. It's just the slope of that line, which you correctly found. Even though the velocity is decreasing at t = 35, the velocity is still positive. If you wanted to use the -50 m/s value, you would have to use the time interval 15 second time interval from 100 m/s to -50 m/s. I misread that the first time I looked at it too.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Graph and Questions
  1. Graph Question (Replies: 1)

  2. Graph Questions (Replies: 3)

  3. Graph Question (Replies: 1)

  4. Graph Question (Replies: 1)

  5. Graph question (Replies: 1)

Loading...