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RavenLynette
Group Title
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
 one year ago
RavenLynette Group Title
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
 one year ago

This Question is Closed

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

RavenLynette Group TitleBest ResponseYou've already chosen the best response.0
no :/
 one year ago

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

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

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

RavenLynette Group TitleBest ResponseYou've already chosen the best response.0
5.1108 but what's the equation of motion
 one year ago

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

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

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

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

ybarrap Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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