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Rohangrr

  • 2 years ago

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

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  1. baddinlol
    • 2 years ago
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    Does it start from rest?

  2. Rohangrr
    • 2 years ago
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    |dw:1342006345931:dw|

  3. Rohangrr
    • 2 years ago
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    thanks

  4. baddinlol
    • 2 years ago
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    Wait

  5. baddinlol
    • 2 years ago
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    Its wrong

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

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

  8. baddinlol
    • 2 years ago
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    Oops sorry

  9. Rohangrr
    • 2 years ago
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    lol yes firstly u found the mistake

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

  11. Rohangrr
    • 2 years ago
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    @Shane_B can u make it more clearer

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

  13. Shane_B
    • 2 years ago
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    Which part isn't clear?

  14. Rohangrr
    • 2 years ago
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    well plzzz make it clear from the beggining

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

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

  17. baddinlol
    • 2 years ago
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    t = 10s

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

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

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

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

  22. Rohangrr
    • 2 years ago
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    not 40 m/s2

  23. baddinlol
    • 2 years ago
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    no no

  24. baddinlol
    • 2 years ago
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    The final velocity is 40m/s

  25. Shane_B
    • 2 years ago
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    40m/s is the final velocity...

  26. Rohangrr
    • 2 years ago
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    okay lol

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