So at time zero a particle is at x= 4 m and y= 3 m and has a velocity of $${ v= \left( 2.0 \ \boldsymbol{\hat{\imath}}-9.0 \ \boldsymbol{\hat{\jmath}} \right) \text{m/s}}$$
The acceleration of the particle is constant and given by
$${a= \left(4.0 \ \boldsymbol{\hat{\imath}}+3.0 \ \boldsymbol{\hat{\jmath}} \right) \text{m/s}^2}$$
Express the position vector at t= 4.0 sec in terms of $\boldsymbol{\hat{\imath}}$ and $\boldsymbol{\hat{\jmath}}$. Also give the magnitude and direction of the position vector at this time.
I believe I have to use the velocity formula and divide the two numbers by ${t= 4}$ sec. From there I think I have the $x$ and add it to the x and y when ${t= 0}$ sec. But I don't know what to do from there.
Best Answer
Formally you would have to integrate the acceleration in order to find the velocity, and then integrate the velocity to find the position as a function of time. But given that this is a uniformly accelerated motion, the solution is well known, and of the form $$\mathbf x = \mathbf x_0 +\mathbf v_0t + \frac12\mathbf at^2,$$ which is just a special case of the more general procedure that I have mentioned above.