Particle acceleration/displacement question

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The discussion focuses on a particle's motion in the x-y plane, starting at the position vector 6i - 2j. The particle's velocity is defined as 3cos(3t)i - 2sin(6t)j, leading to the derived acceleration of -9sin(3t)i - 12cos(6t)j. The initial position is clarified as the starting point, not the displacement, with the correct displacement being zero at t=0. The importance of understanding ijk notation for vector equations is emphasized, as it aids in managing components effectively. Resources and hints for mastering ijk notation in mechanics are encouraged for better comprehension.
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Homework Statement


A particle moves in the x-y plane relative to a fixed point O. Initially, the particle is at the point 6i-2j, where I and J are position vectors in the directions of the x and y-axis respectively. The particle moves such that t seconds after the start of its moition, the velocity of the particle is given by 3cos(3t)i - 2sin(6t)j. Find an expression in terms of t for:

a) acceleration of particle
b) its displacement
[Hint: Use radiains]

Homework Equations


a) Chain rule for differenting dy/dx = dt/dx * dt/dx the function 3cos(3t)i - 2sin(6t)j
b) Integration of function 3cos(3t)i - 2sin(6t)j

The Attempt at a Solution


a) -9sin(3t)i - 12cos(6t)j
b) sin(3t)i + 1/3cos(6t)j + c.

Further
I don't really get ijk notation or how to use it effectively. I know how to switch from polar to component form for simple equations, but with trigonometric functions I'm getting really confused. I have no idea if what I'm doing is correct for this. I'm assuming 6i-2j is its displacement, but I'm not sure what to do with it. I know t= d/v, but that's about it. Any hints/resources to help me get around using ijk in mechanics based questions would be great.

Edit: I realized that if t=0, then s = 6i-2j. So to solve for c, 6i-2j = sin(3t)i + 1/3cos(6t)j + c. But aye, any useful redirects to worthwhile resources on ijk stuff would be great.
 
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Try to embrace and understand the ijk notation instead of getting around it. It helps you write and manipulate vector equations keeping the x, y and z components sorted out. Your assumption that ##6\hat i-2\hat j## is the displacement is incorrect. The problem clearly states that it is the initial position of the particle. It's the position where you will find the particle at t = 0. This means that if you replace ##t## with ##0## in your expression for the position (not the displacement), you should get ##6\hat i-2\hat j##, but if you replace ##t## with ##0## in your expression for the displacement, you should get zero.

It seems that so far you have used the unit vector as effectively as is possible.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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