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anonymous

  • 5 years ago

A particle moves under a hypothetical force so that its velocity vector is v= <ktcost, ktsint>. Find its radial acceleration...Please help...

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  1. anonymous
    • 5 years ago
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    whata hard bout that?

  2. anonymous
    • 5 years ago
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    acceleration is the derivative of velocity with respect to time

  3. anonymous
    • 5 years ago
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    assuming k is a constant

  4. anonymous
    • 5 years ago
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    \[a = < k \frac{d}{dt} ( tcos(t)) , k \frac{d}{dt} ( tsin(t) ) > \]

  5. anonymous
    • 5 years ago
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    k can be factored out of the vector, and it will just provide a scalar multiple

  6. anonymous
    • 5 years ago
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    \[a = k < ( \cos(t) -tsin(t)) , (\sin(t) + tcos(t)) > \]

  7. anonymous
    • 5 years ago
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    by product rules

  8. anonymous
    • 5 years ago
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    wow you are such a genius. I admire you all

  9. anonymous
    • 5 years ago
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    ?

  10. anonymous
    • 5 years ago
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    bt i need the radial accn, not the total accn....

  11. anonymous
    • 5 years ago
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    at time t=0

  12. anonymous
    • 5 years ago
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    yeh i dnt have clue on that :|

  13. anonymous
    • 5 years ago
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    radial is apparently equal to centripetal= v^2 / r

  14. anonymous
    • 5 years ago
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    but i dont have a clue

  15. anonymous
    • 5 years ago
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    hey i got it neway ..thnx

  16. nikvist
    • 5 years ago
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    \[\vec{a}_n=<-kt\sin{t},kt\cos{t}>\]

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