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RavenLynette

  • 2 years ago

A particle at rest undergoes an acceleration of 2.6 m/s^2 to the right and 4.4 m/s^2 up What is its speed after 7.4 s?

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  1. shubham.bagrecha
    • 2 years ago
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    you know vectors?

  2. RavenLynette
    • 2 years ago
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    no :/

  3. shubham.bagrecha
    • 2 years ago
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    |dw:1379213971917:dw|

  4. shubham.bagrecha
    • 2 years ago
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    what's the answer?

  5. ybarrap
    • 2 years ago
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    compute the magnitude of the vector (i.e. Pythagoras theorem) as suggested by @shubham.bagrecha . Then use your equations of motion to determine speed using this magnitude.

  6. shubham.bagrecha
    • 2 years ago
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    net acceration ,\[a=\sqrt{4.4^2+2.6^2}\]

  7. RavenLynette
    • 2 years ago
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    5.1108 but what's the equation of motion

  8. shubham.bagrecha
    • 2 years ago
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    then use, v=u+at

  9. ybarrap
    • 2 years ago
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    u is initial speed (0)

  10. RavenLynette
    • 2 years ago
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    I got 37.81 but that seems like a lot?

  11. ybarrap
    • 2 years ago
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    5 m/s^2 is about 1/2 g. If you fell for 7 seconds at 1 g your speed would be 72 m/s, so 37.8 doesn't seem to far off what you would expect given that acceleration.

  12. RavenLynette
    • 2 years ago
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    Oh wow well thanks :) could I use cosine to find the direction between -180 and 180 degrees?

  13. ybarrap
    • 2 years ago
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    You could use \(\arctan(4.4/2.6)\) to find the angle between the direction of motion and the horizontal, which in this case is about 60\(^\circ\). But, it's not necessary to solve this problem.

  14. RavenLynette
    • 2 years ago
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    Thanks for your help :)

  15. ybarrap
    • 2 years ago
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    yeah, np

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