# Particle acceleration/displacement question

## 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

## 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 realised 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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kuruman
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.