Source: Principles of Physics
Problem Sets:
Problem
A particle moves in a plane with acceleration given by $\vec{a}(t) = (k_x t)\hat{i} + (k_y t)\hat{j}$, where $k_x$ and $k_y$ are constants. At $t=0$, its initial position is $\vec{r}_0 = r_{0x}\hat{i} + r_{0y}\hat{j}$ and its initial velocity is $\vec{v}_0 = v_{0x}\hat{i} + v_{0y}\hat{j}$.
- What is its position vector $\vec{r}(t)$ at time $t$?
- What is the angle $\phi(t)$ its velocity vector makes with the positive x-axis at time $t$?
[Q1] $\vec{r}(t) = \left(r_{0x} + v_{0x}t + \frac{1}{6} k_x t^3\right)\hat{i} + \left(r_{0y} + v_{0y}t + \frac{1}{6} k_y t^3\right)\hat{j}$ [Q2] $\phi(t) = \arctan\left(\frac{v_{0y} + \frac{1}{2} k_y t^2}{v_{0x} + \frac{1}{2} k_x t^2}\right)$
The velocity vector $\vec{v}(t)$ is found by integrating the acceleration vector $\vec{a}(t)$ with respect to time, starting from the initial velocity $\vec{v}_0$.
$$ \vec{v}(t) = \vec{v}_0 + \int_0^t \vec{a}(t') dt' = (v_{0x}\hat{i} + v_{0y}\hat{j}) + \int_0^t ((k_x t')\hat{i} + (k_y t')\hat{j}) dt' $$ $$ \vec{v}(t) = \left(v_{0x} + \frac{1}{2}k_x t^2\right)\hat{i} + \left(v_{0y} + \frac{1}{2}k_y t^2\right)\hat{j} $$The position vector $\vec{r}(t)$ is found by integrating the velocity vector $\vec{v}(t)$, starting from the initial position $\vec{r}_0$.
$$ \vec{r}(t) = \vec{r}_0 + \int_0^t \vec{v}(t') dt' = (r_{0x}\hat{i} + r_{0y}\hat{j}) + \int_0^t \left[\left(v_{0x} + \frac{1}{2}k_x (t')^2\right)\hat{i} + \left(v_{0y} + \frac{1}{2}k_y (t')^2\right)\hat{j}\right] dt' $$ $$ \vec{r}(t) = \left(r_{0x} + v_{0x}t + \frac{1}{6}k_x t^3\right)\hat{i} + \left(r_{0y} + v_{0y}t + \frac{1}{6}k_y t^3\right)\hat{j} $$The angle of the velocity vector is found from the ratio of its components, $v_y(t)$ and $v_x(t)$.
$$ \phi(t) = \arctan\left(\frac{v_y(t)}{v_x(t)}\right) = \arctan\left(\frac{v_{0y} + \frac{1}{2} k_y t^2}{v_{0x} + \frac{1}{2} k_x t^2}\right) $$