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1 d kinematics

  1. Feb 1, 2009 #1
    1. why do the units for time appear twice in the units for acceleration
    2. a = v/t - v1-vo/t1-t0
    3. once for speed change and once for time change.

    1. what is the acceleration of a car that travels in a straight line at a constant speed of 27 m/s?
    2. 27m/s / 10s
    3. 2.7m
  2. jcsd
  3. Feb 1, 2009 #2


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    Is this:

    1. why do the units for time appear twice in the units for acceleration

    your first question?

    Is this:

    2. a = v/t - v1-vo/t1-t0

    a possible answer?

    It's not clear what you would like help with. Identify your question, then show your attempt to answer it.
  4. Feb 1, 2009 #3
    a = v/t = v1-v0/t1-t0

    i just want to know if i got the right answer or not
  5. Feb 1, 2009 #4
    [tex] a=\frac{\Delta v}{\Delta t}= \frac{v_{f}-v_{o}}{t_{f}-t_{o}} [/tex]

    you can't write it like "v1-v0/t1-t0" this means that v1 minus (vo over t1) minus t0.
  6. Feb 1, 2009 #5

    yeah thats the equation i thought would relate, but are my answer correct?
  7. Feb 1, 2009 #6
    What exactly is your question?
  8. Feb 1, 2009 #7
    both number 1
  9. Feb 2, 2009 #8


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    Okay, now that you have told us that Number 1 is the question in both cases, we can help you. First, your answer to question 1 is not correct. Think about the definitions. Velocity is change in position divided by change in time so the units are length/time or

    [tex] V = \frac{L}{T} [/tex]

    Now acceleration is change in velocity divided by time. Take the equation I just wrote and do the algebra. What do you see?

    For the second question. What is the adjective modifying the word speed??? What is the definition of velocity that I just wrote? Put those tow ideas together and what do you get?

    Hint: With physics sometime it's useful to ask "What does it mean?" and then go back and look at the definitions.
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