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Velocity, Speed, Acceleration, and CAPA homework problem - I'm stumped!

  1. Jun 5, 2009 #1
    1. The problem statement, all variables and given/known data

    "A car travels a certain distance along a straight road and on the way must stop at traffic lights and obey local speed limits. The figure (not pictured but info filled in below) shows the distance the car travels as a function of time. Choose all the correct answers which apply to the speed of the car.
    A. The speed at 2.1min (distance = 0.35miles) is less than at 2.4min (distance = 0.55miles)
    B. At 2.3min (distance = 0.5miles) the speed is as high as it gets.
    C. The speed at 6.0min (distance = 1.30miles) is less than at 2.3min (dist = 0.5miles)
    D. The speed is zero at 0.5min (dist = 0miles straight line on graph) and at 3.7min (distance =1.05miles but straight line on graph)
    E. The speed does not change from 3min (dist. = 0.95miles) to 3.7min (dist = 1.05miles)

    2. Relevant equations
    Am I incorrect in my approach with the s=d/t equation? Should my approach be simply taking the magnitude of the velocity vector? Along that line, I am confused regarding "Speed" "Velocity" and "Acceleration" in their relations and definitions and how to apply this to the above and other problems.


    3. The attempt at a solution
    I have assumed speed s=d/t and have converted the units, although I don't think that is necessary, and I assumed the following were correct at multiple tries (A, AC, AD, ADE) and none of these answers have been correct.

    Here is a link to the above mentioned graph. I couldn't figure the photobucket method. Apologies, but I appreciate the help!

    http://www.flickr.com/photos/mountainmamas/3599718176/?addedcomment=1#comment72157619224566381 [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 5, 2009 #2
    Hello, and welcome to Physics Forums!

    It would be to your benefit to attach an image of the graph you're looking at, or find a link to a website with the graph. I know many people will upload pictures to photobucket. In doing this, you increase the likelihood that someone will be able to respond to your query, and do so accurately.

    Here's some helpful information about your problem:

    Displacement ([tex]\Delta[/tex]x) of an object is its change in position. A displacement is NOT a distance. Distance describes the total units traveled. For example, if you run around a track once, you begin and end at the same spot. On a standard track, the distance you have traveled is 400 m. However, your displacement is 0. You are in the exact same spot that you started.

    As with displacement and distance, speed and velocity are also NOT interchangeable. Velocity is a vector -- meaning that it implies magnitude and direction. On a distance-time graph, average speed would be the total distance over total time. Instantaneous speed would be the slope at a point on the graph, or taking the limit as [tex]\Delta[/tex]t --> 0 of [tex]\Delta[/tex]x/[tex]\Delta[/tex]t.

    Acceleration is the change in velocity over time, or a change in the direction of motion of an object. When the object's velocity and acceleration are in the same direction, the speed of the object is increasing with time. When speed and velocity are in opposite directions, the speed of the object decreases with time.

    On a velocity-time graph, the instantaneous acceleration of an object at a given time equals the slope of the tangent at the point of interest.

    Hope this helps!
     
  4. Jun 5, 2009 #3
    Okay so the image upload did not work. Here goes again. Below is a link to the image for the above question. I have a Flickr account and not a photobucket account, so might as well use it as I'm paying for it!!

    http://www.flickr.com/photos/mountainmamas/3599718176/?addedcomment=1#comment72157619224566381 [Broken]

    http://www.flickr.com/photos/mountainmamas/3599718176/?addedcomment=1#comment72157619224566381 [Broken]
     
    Last edited by a moderator: May 4, 2017
  5. Jun 5, 2009 #4
    [/URL]

    Ah, much better! Ok, what you're looking at is comparing the slope of the line at different times. (Remember: The slope of the line = the speed of the car at that time.)

    Let's look at (A):
    Is the slope at 2.1 min greater or less than the slope at 2.4 min? Is the line "steeper" or not?
     
    Last edited by a moderator: May 4, 2017
  6. Jun 5, 2009 #5
    If I just look at the graph, it appears that the slope at 2.1 and 2.4 is equal. If I were to take that entire time period (from 1.4min to 3.3min) and found the tangent of that line (aka the slope) wouldn't it be equal to the instantaneous velocity at both time 2.1min and 2.4min?

    I hope that made sense... if not, let me know. Thank you so much for helping me out!
     
    Last edited: Jun 5, 2009
  7. Jun 5, 2009 #6
    Nice job. What does that tell you about (A)?
     
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