Finding distance from origin using acceleration

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SUMMARY

The particle's acceleration is defined by the equation a(t) = 3ti + 4tj, where a is in m/s² and t is in seconds. To determine the distance from the origin at t = 2 seconds, the velocity is first calculated by integrating the acceleration, resulting in v(t) = (3t²)/2 i + 2t² j. The distance is then found by integrating the velocity, yielding x(t) = (t³)/2 i + (2t³/3) j. The correct distance, calculated using the magnitude of the vector at t = 2 seconds, is approximately 6.67 meters, aligning closely with the teacher's estimate of 7 meters.

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Entr0py
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Homework Statement


Acceleration of a particle that begins at rest at the origin is given by a(t)=3ti+4tj, where a is in m/s^2 and t is in seconds. The particle's distance from the origin at time t=2s is what?

Homework Equations


You need to find velocity then distance

The Attempt at a Solution


To find velocity I integrated the acceleration (I haven't covered integration in calculus yet, so it's a bit difficult to do). I got v(t)=(3t^2)/2i+2t^2j. Now to find distance I integrate velocity. I got x(t)=(t^3)/2i+(2t^3/3)j. Plugging in t=2 s, I get x=9.33 m. But my teacher says the answer is about 7 m.
 
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Entr0py said:
To find velocity I integrated the acceleration (I haven't covered integration in calculus yet, so it's a bit difficult to do). I got v(t)=(3t^2)/2i+2t^2j. Now to find distance I integrate velocity. I got x(t)=(t^3)/2i+(2t^3/3)j. Plugging in t=2 s, I get x=9.33 m. But my teacher says the answer is about 7 m.
You did fine with your integrations! But note that the result is a vector value with i and j components (or x and y components if you wish). How do you find the magnitude of a vector?
 
You would do the square root of the i and j hats right?
 
gneill said:
You did fine with your integrations! But note that the result is a vector value with i and j components (or x and y components if you wish). How do you find the magnitude of a vector?
And thank you for your response.
 
Entr0py said:
You would do the square root of the i and j hats right?
Square root of the sum of the squares. Like finding the hypotenuse of a right angle triangle. This is also called "adding in quadrature" if you're interested in the lingo.
 
Thank you man. I found the distance by finding the i and j hat separately. I got 4ti+(16/3)tj m. I plus in t=2 s and I get square root of 44.4. which is about 6.67 which is close to 7 m. Thanks a lot for helping me.
 
gneill said:
Square root of the sum of the squares. Like finding the hypotenuse of a right angle triangle. This is also called "adding in quadrature" if you're interested in the lingo.
You already know I'm interested in the lingo.
 

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