Position vector R of a particle moving x-y plane

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SUMMARY

The position vector R of a particle moving in the x-y plane is defined as r = (2t^3)i + (6-7t^4)j, where r is in meters and t is in seconds. To calculate the position, velocity, and acceleration at t = 2 seconds, one must substitute t into the equation for r to find the displacement. The velocity v and acceleration a can be determined by taking the first and second derivatives of the position vector, respectively. This method is essential for accurately solving problems related to motion in physics.

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  • Knowledge of calculus, specifically differentiation
  • Familiarity with kinematic equations in physics
  • Basic grasp of motion in two dimensions
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  • Study the application of kinematic equations in two-dimensional motion
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chris61986
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Hello, new to these forums! I'm working on my UPI homework, and I just want to verify if I'm doing something correctly. The problem:

The position vector r of a particle moving in the x - y plane given by r = (2t^3)i + (6-7t^4)j, where r is in meters and t is in seconds.
Calculate
r
v
a
when t = 2s

Now, I'm pretty sure I know how to do it, but I don't get a second chance once I turn the homework in! :)

So in the original equation, r is meters. So I'm thinking this is the displacement.
If this is the displacement, I just plug in 2 for t and solve.
Because that's the displacement, I can find a and v from the derivatives of the original function.

Is this correct?
 
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Sounds good. Just remember that [itex]\vec r[/itex] is a vector.
 

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