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ksaimouli

  • 3 years ago

The position of a particle moving along a line is given by

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  1. ksaimouli
    • 3 years ago
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    \[s(t)=2t^3-24t^2+90t+7\]

  2. ksaimouli
    • 3 years ago
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    for what values t is the speed of particle increasing

  3. Jonask
    • 3 years ago
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    \[s'(t) \implies velocity \space \space s''(t) \implies acceleration\]

  4. ksaimouli
    • 3 years ago
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    ya i got a(t)=0 i got 4

  5. ksaimouli
    • 3 years ago
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    as critical point

  6. ceb105
    • 3 years ago
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    ds/dt=6t^2-48t+90 thus d^2s/dt^2=12t-48 => aceleration is greater than zero when t>4

  7. Jonask
    • 3 years ago
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    well first get the turning point, you mean s'(t)=0

  8. ksaimouli
    • 3 years ago
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    no i mean s''(t) gives acceleration =0 will give us information about velocity

  9. Jonask
    • 3 years ago
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    yes so you got 4 right

  10. ksaimouli
    • 3 years ago
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    yes

  11. ceb105
    • 3 years ago
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    So yea it is when t>4 that the acceleration is postivei .e. speed increases

  12. Jonask
    • 3 years ago
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    yes

  13. ksaimouli
    • 3 years ago
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    no that not the answer i am looking for

  14. ceb105
    • 3 years ago
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    Well its the right one...

  15. ksaimouli
    • 3 years ago
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    but the answer is 3<t<4 and t>5 ( i was shocked !!!!

  16. ceb105
    • 3 years ago
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    is it possible that this is wrong?

  17. ksaimouli
    • 3 years ago
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    @hartnn

  18. Jonask
    • 3 years ago
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    |dw:1360515901447:dw| graph increasing the speed inceases when the grapg rises so x<a and x>b

  19. ceb105
    • 3 years ago
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    Okay well 3 and 5 are the critical points of the velocity

  20. Jonask
    • 3 years ago
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    |dw:1360516136831:dw|

  21. Jonask
    • 3 years ago
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    |dw:1360516160914:dw|

  22. ksaimouli
    • 3 years ago
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    But why i need the critical points of velocity when they ask for velocity

  23. ksaimouli
    • 3 years ago
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    This is no calculator question so i cannot graph

  24. Jonask
    • 3 years ago
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    they did not say the rate of change in velocity nor speed just velocity increasing

  25. Jonask
    • 3 years ago
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    yeah i know what i am saying is\[s'(t)=0\]

  26. ksaimouli
    • 3 years ago
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    i e 0,3,5

  27. ksaimouli
    • 3 years ago
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    then how did they get 4

  28. Jonask
    • 3 years ago
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    so you get 3 and 5 right

  29. ksaimouli
    • 3 years ago
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    yes

  30. Jonask
    • 3 years ago
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    speed =|velocity| \[|6t^2-8t+15|>0\] i would like to explain grapghically 4 a sec

  31. Jonask
    • 3 years ago
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    i mean\[|t^2-8t+15|\]

  32. ksaimouli
    • 3 years ago
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    its actually velocity=I speedI

  33. ksaimouli
    • 3 years ago
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    velocity is vector where as speed is scalar

  34. Jonask
    • 3 years ago
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    i agree with person 3

  35. ksaimouli
    • 3 years ago
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    i found out that they used velocity and acceleration derivatives and they set them =0 and found the answer but i have learned that to find graph of y' use y'' to find max or min

  36. Jonask
    • 3 years ago
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    yes so we found \[|t^2-8t+15|>0 \implies t<3 ,t>5\] but also for - velocity we put \[-(t^2-8t+15)\] its positive for \[-(2t-8)>0 \implies t<4\] (3,4) and x>5

  37. Jonask
    • 3 years ago
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    did the link help

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