anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
you know vectors?
anonymous
  • anonymous
no :/
anonymous
  • anonymous
|dw:1379213971917:dw|

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anonymous
  • anonymous
what's the answer?
ybarrap
  • ybarrap
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.
anonymous
  • anonymous
net acceration ,\[a=\sqrt{4.4^2+2.6^2}\]
anonymous
  • anonymous
5.1108 but what's the equation of motion
anonymous
  • anonymous
then use, v=u+at
ybarrap
  • ybarrap
u is initial speed (0)
anonymous
  • anonymous
I got 37.81 but that seems like a lot?
ybarrap
  • ybarrap
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.
anonymous
  • anonymous
Oh wow well thanks :) could I use cosine to find the direction between -180 and 180 degrees?
ybarrap
  • ybarrap
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.
anonymous
  • anonymous
Thanks for your help :)
ybarrap
  • ybarrap
yeah, np

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