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Rohangrr

okay @lgbasallote @Shane_B @baddinlol A racing car has a uniform acc.. of \[ 4 \quad m/s^2\]. What's the distance she will cover 10s later

  • one year ago
  • one year ago

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  1. baddinlol
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    Does it start from rest?

    • one year ago
  2. Rohangrr
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    |dw:1342006345931:dw|

    • one year ago
  3. Rohangrr
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    thanks

    • one year ago
  4. baddinlol
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    Wait

    • one year ago
  5. baddinlol
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    Its wrong

    • one year ago
  6. baddinlol
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    Use the formula v = ut + 0.5at^2

    • one year ago
  7. Shane_B
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    Assuming a start from rest: \[d=Vi(t)+\frac{1}{2}at^2=0m/s^2+\frac{1}{2}(4m/s^2)(10s)^2=200m\]

    • one year ago
  8. baddinlol
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    Oops sorry

    • one year ago
  9. Rohangrr
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    lol yes firstly u found the mistake

    • one year ago
  10. baddinlol
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    x = ut + 0.5at^2

    • one year ago
  11. Rohangrr
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    @Shane_B can u make it more clearer

    • one year ago
  12. Shane_B
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    Or, you can use: \[d=\frac{V_i+V_f}{2}t=\frac{0m/s+40m/s}{2}(10s)=200m\]

    • one year ago
  13. Shane_B
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    Which part isn't clear?

    • one year ago
  14. Rohangrr
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    well plzzz make it clear from the beggining

    • one year ago
  15. baddinlol
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    Well he used this formula x = ut + 0.5at^2

    • one year ago
  16. baddinlol
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    u = 0m/s , becausehe assumed the car started from rest

    • one year ago
  17. baddinlol
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    t = 10s

    • one year ago
  18. baddinlol
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    a = 4m/s^2

    • one year ago
  19. baddinlol
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    Then he plugged in the numbers x = 0.5*4* 100 = 200m

    • one year ago
  20. Shane_B
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    @baddinlol: thanks...working with someone else at the moment :)

    • one year ago
  21. Rohangrr
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    @Shane_B ITS \[\huge 4 \quad m/s^2 \]

    • one year ago
  22. Rohangrr
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    not 40 m/s2

    • one year ago
  23. baddinlol
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    no no

    • one year ago
  24. baddinlol
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    The final velocity is 40m/s

    • one year ago
  25. Shane_B
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    40m/s is the final velocity...

    • one year ago
  26. Rohangrr
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    okay lol

    • one year ago
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