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rvc

  • one year ago

The motion of a particle along a straight line is governed by the relation a= t^3-2t+7 where a is in m/s^2 and t is time in seconds. At time t=1 the velocity of the particle is 3.58m/s and displacement is 9.39m. Calculate the displacement,velocity and acc at t=2s. I know the formula but the calculation part is getting wrong somewhere.

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  1. Michele_Laino
    • one year ago
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    you have to integrate that function in order to find the velocity v(t): \[\Large v\left( t \right) = \int {\left( {{t^3} - 2t + 7} \right)dt} \]

  2. rvc
    • one year ago
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    v = \[\rm v=\frac{ t^4 }{ 4 } -2\frac{ t^3 }{ 3 }+7t+C\]

  3. rvc
    • one year ago
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    brb

  4. Michele_Laino
    • one year ago
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    ok! now you have to substitute your condition, namely t=1 and v=3.58, so you will find the value of the constant C

  5. Michele_Laino
    • one year ago
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    I got this: \[\Large v\left( t \right) = \int {\left( {{t^3} - 2t + 7} \right)dt} = \frac{{{t^4}}}{4} - {t^2} + 7t + C\]

  6. rvc
    • one year ago
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    yes

  7. Michele_Laino
    • one year ago
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    after a substitution, you will get this: \[\Large 3.58 = \frac{1}{4} - 1 + 7 + C\]

  8. rvc
    • one year ago
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    \[\sf 3.58=\frac{ 1 }{ 4 }-\frac{ 2 }{ 3 } +7+C\]

  9. Michele_Laino
    • one year ago
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    why 2/3?

  10. rvc
    • one year ago
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    OHHHHHHHHHHHHHHHHHHhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh i got it correctly now because i copied the question wrong

  11. Michele_Laino
    • one year ago
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    is a(t) like this: \[\Large a\left( t \right) = {t^3} - 2t + 7\]

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