Finding distance from origin using acceleration

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Homework Help Overview

The problem involves determining the distance of a particle from the origin given its acceleration function, a(t) = 3ti + 4tj, where the particle starts from rest. The discussion centers around finding the distance at t = 2 seconds.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of acceleration to find velocity and then distance. There are questions about how to calculate the magnitude of the resulting vector and the proper interpretation of the components.

Discussion Status

Some participants have provided guidance on calculating the magnitude of the vector, while others have shared their attempts and results. There is an ongoing exploration of the calculations and interpretations without a clear consensus on the final answer.

Contextual Notes

One participant notes difficulty with integration due to a lack of coverage in their calculus studies. There is also mention of differing answers from the teacher, which raises questions about the assumptions made in the calculations.

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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