# Thread: Mechanics with vectors question!

1. ## Mechanics with vectors question!

Alright all ive been attempting this question and ive become stuck.

This is the question.

So for the first part i wrote F=Mx^.. (X double dot).
I also stated that if there was no force acting on the particle then the particle would stay at a constant speed V if we ignored instances such as the particle being on a slope.

For the second part i wrote (Im using X to denote x hat (x^^), same applied with y.)

F_x X + F_y Y = m x^.. X + m y^.. Y

=> x^.. = 2 , y^.. = 3x^2

Now for the third part when writing down the inital conditions this is where i become confused. Shall i write it in plain english such as;

At time equal to 0, this displacement of the particle x^.. will be 0.

Im not too sure how to answer that part, or the last part.

Any help is greatly appricated cheers people

2. Hi

Integrating $\displaystyle \frac{d^2x}{dt^2} = 2$ leads to $\displaystyle \frac{dx}{dt} = 2t + C$

The particle starts at rest at time t=0. This means that $\displaystyle \frac{dx}{dt} (t=0) = 0$ and therefore C=0 and $\displaystyle \frac{dx}{dt} = 2t$

Integrating $\displaystyle \frac{dx}{dt} = 2t$ leads to $\displaystyle x = t^2 + K$

The particle starts at the origin at t=0. This means that $\displaystyle x(t=0) = 0$ and therefore K=0 and $\displaystyle x = t^2$

Now substitute $\displaystyle x = t^2$ into $\displaystyle \frac{d^2y}{dt^2} = 3x^2$ and solve the same way