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roadjester

  • 2 years ago

An object moves with constant acceleration 4.00 m/s^2 and over a time interval reaches a final velocity of 12.0 m/s. a) If its initial velocity is -6.00 m/s, what is its displacement during the time interval. b) What is the total distance it travels during the time interval in part b?

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  1. alphyshaju
    • 2 years ago
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    0

  2. lucifur
    • 2 years ago
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    specify what part b of the motion is?

  3. roadjester
    • 2 years ago
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    oops, it's supposed to be "What is the total distance it travels during the time interval in part a" not part b. Typo.

  4. lucifur
    • 2 years ago
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    27/2 and total distance equals 45/2

  5. roadjester
    • 2 years ago
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    Uh, how do you figure? I mean, what formula did you use? (kinematics)

  6. lucifur
    • 2 years ago
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    \[v ^{2}-u ^{2} = 2aS\] v=12,u=-6 and a=4 u get total displacement S..

  7. lucifur
    • 2 years ago
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    for, total distance use the same kinematic equation but twice once till the velocity becomes zero thats from u=-6 to v=0; then again from u=0 to v=12

  8. roadjester
    • 2 years ago
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    I'm guessing u=initial velocity and v=final velocity. I get v, but why u?

  9. roadjester
    • 2 years ago
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    And why stop at 0? I mean, I can't visualize the two parts but I think I get the concept.

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