Variable Acceleration with Vectors (Mechanics)
Acceleration of a particle using Vectors
The acceleration tells you how the velocity, $\mathbf{v}$ changes with time, $t$. We can use the relationship $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ to solve certain problems, where $\mathbf{u}$ represents initial velocity.
Worked Example
Example
Suppose that a particle has velocity $(-4 \mathbf{i} + 3\mathbf{j}) \mathrm{ms^{-1} }$ at time $t=0$. The particle moves with constant acceleration $\mathbf{a} = (-2\mathbf{i} + \mathbf{j}) \mathrm{ms^{-2} }$. Find the speed of the particle after $4 \mathrm{seconds}$.
Solution
We can use the formula $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ to find the velocity of the particle, and from this find the speed. \begin{align} \mathbf{v} & = \mathbf{u} + \mathbf{a}t, \\ & = (-4 \mathbf{i} + 3 \mathbf{j}) + \left( (-2 \mathbf{i} + \mathbf{j}) \times 4 \right), \\ & = (-4 \mathbf{i} + 3 \mathbf{j}) + (-8 \mathbf{i} + 4\mathbf{j}), \\ & = (-4 - 8) \mathbf{i} + (3 + 4) \mathbf{j}, \\ & = (-12 \mathbf{i} + 7 \mathbf{j} ) \mathrm{ms^{-1} }. \end{align} Then we take the magnitude to find the speed of the particle. \begin{align} \text{speed} & = \lvert \mathbf{v} \rvert, \\ & = \sqrt{ 12^2 + 7^2}, \\ & = \sqrt{193}, \\ & = 13.9 \mathrm{ms^{-1} } \text{ (to 3s.f.)} \end{align}
Integration and Differentiation
The relationship between displacement, $\mathbf{r}$, velocity, $\mathbf{v}$ and acceleration, $\mathbf{a}$ is the same when the quantities are represented as vectors or as scalars.
We can write the vector $\mathbf{r}$ as \[\mathbf{r} = x\mathbf{i} + y\mathbf{j}.\] Where $\mathbf{i},\mathbf{j}$ are the unit vectors in the $x$ and $y$ directions respectively.
Therefore we can write velocity and acceleration as \begin{align} \mathbf{v} = \dot{\mathbf{r}} &= \dot{x}\mathbf{i} + \dot{y}\mathbf{j},\\ \mathbf{a} = \ddot{\mathbf{r}} &= \ddot{x}\mathbf{i} + \ddot{y}\mathbf{j}, \end{align} where the dot above $x$ denotes the derivative of $x$.
Worked Example
Example
A particle $P$ is moving in a plane. At time $t$ seconds, its velocity $\mathbf{v}\mathrm{ms^{-1} }$ is given by \[\mathbf{v} = 4t\mathbf{i} + \dfrac{3}{2}t^2\mathbf{j}.\] When $t=0$, the position vector of $P$ with respect to the origin $O$ is $3\mathbf{i} - 4 \mathbf{j}$. Find the position vector of $P$ at time $t$ seconds.
Solution
To find the position vector, $\mathbf{r}$, we need to integrate $\mathbf{v}$ with respect to $t$. \begin{align} \mathbf{r} &= \int \mathbf{v} \mathrm{d}t,\\ &= \int \left(4t\mathbf{i} + \dfrac{3}{2}t^2\mathbf{j}\right) \mathrm{d}t,\\ &= 2t^2\mathbf{i} + \dfrac{1}{2}t^3\mathbf{j} + \mathbf{c}. \end{align} To find the value of $\mathbf{c}$ we need to use the fact that at $t=0,$ we have $\mathbf{r} = 3\mathbf{i} - 4 \mathbf{j}$. Substituting these values into the above equation gives \begin{align} 3\mathbf{i} - 4 \mathbf{j} &= 2\times 0 \mathbf{i} + \dfrac{1}{2} \times 0^3\mathbf{j} + \mathbf{c},\\ \Rightarrow 3\mathbf{i} - 4 \mathbf{j} &= \mathbf{c}.\\ \end{align} So the position vector is \begin{align} \mathbf{r} &= 2t^2\mathbf{i} + \dfrac{1}{2}t^3\mathbf{j} + 3\mathbf{i} - 4 \mathbf{j},\\ &= (2t^2+3)\mathbf{i} + \left(\dfrac{1}{2}t^3 - 4\right)\mathbf{j}. \end{align}