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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?
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?

This Question is Closed

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

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

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

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

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

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

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

ybarrap
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for your help :)
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