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 one year 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?
 one year 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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shubham.bagrecha
 one year ago
Best ResponseYou've already chosen the best response.1you know vectors?

shubham.bagrecha
 one year ago
Best ResponseYou've already chosen the best response.1dw:1379213971917:dw

shubham.bagrecha
 one year ago
Best ResponseYou've already chosen the best response.1what's the answer?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0compute 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.

shubham.bagrecha
 one year ago
Best ResponseYou've already chosen the best response.1net acceration ,\[a=\sqrt{4.4^2+2.6^2}\]

RavenLynette
 one year ago
Best ResponseYou've already chosen the best response.05.1108 but what's the equation of motion

shubham.bagrecha
 one year ago
Best ResponseYou've already chosen the best response.1then use, v=u+at

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0u is initial speed (0)

RavenLynette
 one year ago
Best ResponseYou've already chosen the best response.0I got 37.81 but that seems like a lot?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.05 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.

RavenLynette
 one year ago
Best ResponseYou've already chosen the best response.0Oh wow well thanks :) could I use cosine to find the direction between 180 and 180 degrees?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0You 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.

RavenLynette
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your help :)
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