AP physics. one answer to this

Physics
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A particle moving at a velocity of 2.3 m in the positive x direction is given an acceleration of 4.9 m/s^2 in the positive y direction for 4.3s.  What is the speed of the particle? Answer in units of m/s. 

Sep 27th, 2015

Thank you for the opportunity to help you with your question!

if a=4.9m/s^2 after 4.3s the velocity will be (Vf=a x t)  

so 4.9 x 4.3=21.07m/s in y direction and 2.3m/s in x direction

so the total speed is sqrt[(21.07)^2 +(2.3)^2]= 21.2m/s with the angle of arctg (21.07/2.3)= 83 degree

 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 27th, 2015

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Sep 27th, 2015
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Sep 27th, 2015
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