Kinematics: Getting Started

by Keith50, May 13, 2021, 9:50 AM

$\Large \textbf{\color{green}{Kinematics: Getting Started}}$
Hey guys, this marks my first blog on a physics topic, let's get started with kinematics!
$\large \textbf{\color{blue}{(i) Position, Velocity, and Acceleration}}$
In kinematics, we want to describe the motion of a particle and we can do this by using vectors to specify its position, velocity and acceleration.
At any particular time $t$ , the position of the particle with coordinates $(x,y,z)$ in three-dimension can be represented as a $\textit{position vector}$ $\overset{\rightarrow}{\mathbf{r}}$ : $\overset{\rightarrow}{\mathbf{r}}=x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}}$ where $\mathbf{\hat{i}},\mathbf{\hat{j}},$ and $\mathbf{\hat{k}}$ are Cartesian unit vectors.
If the particle is located at position $\overset{\rightarrow}{\mathbf{r_1}}$ at time $t_1$ and it moves along its path to position $\overset{\rightarrow}{\mathbf{r_2}}$ at time $t_2$ , we denote its change in position during that interval by the $\textit{displacement vector}$ $\Delta \overset{\rightarrow}{\mathbf{r}}$ : $\Delta \overset{\rightarrow}{\mathbf{r}}=\overset{\rightarrow}{\mathbf{r_2}} -\overset{\rightarrow}{\mathbf{r_1}}.$ We can then define its $\textit{average velocity}$ $\overset{\rightarrow}{\mathbf{v}_{\textrm{av}}}$ in that interval as the displacement divided by the time interval during which the displacement occurs, $\overset{\rightarrow}{\mathbf{v}_{\textrm{av}}}=\frac{\Delta \overset{\rightarrow}{\mathbf{r}}}{\Delta t}.$ Taking the limit as $\Delta t \rightarrow 0$ , we then have the $\textit{instantaneous velocity}$ $\overset{\rightarrow}{\mathbf{v}}$ as $\overset{\rightarrow}{\mathbf{v}}=\lim_{\Delta t \rightarrow 0} \frac{\Delta \overset{\rightarrow}{\mathbf{r}}}{\Delta t}=\frac{d \overset{\rightarrow}{\mathbf{r}}}{dt}.$ The derivative of the position vector is found by taking the derivative of its components: $\frac{d \overset{\rightarrow}{\mathbf{r}}}{dt}=\frac{d}{dt}(x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}})=\frac{dx}{dt}\mathbf{\hat{i}}+\frac{dy}{dt}\mathbf{\hat{j}}+\frac{dz}{dt}\mathbf{\hat{k}}.$ On the other hand, the velocity vector can also be written in terms of its components $\overset{\rightarrow}{\mathbf{v}}=v_x\mathbf{\hat{i}}+v_y\mathbf{\hat{j}}+v_z\mathbf{\hat{k}}$ where $v_x=\frac{dx}{dt}, \quad v_y=\frac{dy}{dt}, \quad v_z=\frac{dz}{dt}.$ In a similar fashion, we can define the $\textit{average acceleration}$ of the particle in that interval as the change in velocity per time, $\overset{\rightarrow}{\mathbf{a}_{\textrm{av}}}=\frac{\Delta \overset{\rightarrow}{\mathbf{v}}}{\Delta t}.$ The change in velocity means $\overset{\rightarrow}{\mathbf{v}}_{\textrm{final}} -\overset{\rightarrow}{\mathbf{v}}_{\textrm{initial}}$ . By taking the limit as $\Delta t \rightarrow 0$ , we then have the $\textit{instantaneous acceleration}$ $\overset{\rightarrow}{\mathbf{a}}$ as $\overset{\rightarrow}{\mathbf{a}}=\lim_{\Delta t \rightarrow 0} \frac{\Delta \overset{\rightarrow}{\mathbf{v}}}{\Delta t}=\frac{d\overset{\rightarrow}{\mathbf{v}}}{dt}=\frac{d^2\overset{\rightarrow}{\mathbf{r}}}{dt^2}.$ Analoguosly, we can write the instantaneous acceleration vector in terms of its components $\overset{\rightarrow}{\mathbf{a}}=a_x\mathbf{\hat{i}}+a_y\mathbf{\hat{j}}+a_z\mathbf{\hat{k}}$ where $a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt}, \quad a_y=\frac{dv_y}{dt}=\frac{d^2y}{dt}, \quad a_z=\frac{dv_z}{dt}=\frac{d^2z}{dt}.$ $\large \textbf{\color{blue}{(ii) Kinematics Equations for Constant Acceleration}}$
Now, we shall derive several kinematics equations for a particle with constant acceleration. Let's consider a particle moving in one-dimension with a constant acceleration, $a$ , the following equations are true,
1. $\Delta x=\frac{v_{\textrm{initial}}+v_{\textrm{final}}}{2}\Delta t.$ This follows from the graphical interpretation of displacement being the area under a velocity vs. time graph, which is in this case a trapezoid with bases $v_{\textrm{initial}}$ and $v_{\textrm{final}}$ and height $\Delta t$ .
2. $\Delta x=v_{\textrm{initial}}\Delta t+\frac{1}{2}a(\Delta t)^2$ . This again follows from the graphical interpretation of displacement being area under a velocity vs. time graph, but this time, it's a trapezoid with bases $v_{\textrm{initial}}$ and $v_{\textrm{final}}=v_{\textrm{initial}}+at$ , and height $t$ .
3. $\Delta x=v_{\textrm{final}}\Delta t-\frac{1}{2}a(\Delta t)^2$ . This is very similar to the previous case, but now the bases of the trapezoid are $v_{\textrm{final}}-at$ and $v_{\textrm{final}}$ .
4. $v_{\textrm{final}}^2=v_{\textrm{initial}}^2+2a\Delta x$ . This is perhaps the least intuitive of all the kinematics equations, this can be shown true by a simple algebra, as $v_{\textrm{initial}}+v_{\textrm{final}}=2\frac{\Delta x}{\Delta t}$ and $v_{\textrm{initial}}-v_{\textrm{final}}=a\Delta t$ , multiplying both of them yields the result.

Exercise wrote:

For acceleration which is not constant, show that this identity is generalized by the equation $v_{1}^2-v_{0}^2=2\int_{t_0}^{t_1} a \, dx.$ Proof

Now, in general, these kinematics equations holds for a particle moving in $n-$ dimension with constant acceleration, $\overset{\rightarrow}{\mathbf{a}}=\langle a_1,a_2,\ldots a_n\rangle$ as we can decompose its acceleration, velocity and position vectors into their components and work with each component separately. Thus, we still have $\Delta \overset{\rightarrow}{\mathbf{r}}=\frac{\overset{\rightarrow}{\mathbf{v}_1}^2-\overset{\rightarrow}{\mathbf{v_0}}^2}{2}\Delta t$ $\Delta \overset{\rightarrow}{\mathbf{r}}=\overset{\rightarrow}{\mathbf{v}_0}\Delta t+\frac{1}{2}\overset{\rightarrow}{\mathbf{a}}(\Delta t)^2$ $\Delta \overset{\rightarrow}{\mathbf{r}}=\overset{\rightarrow}{\mathbf{v}_1}\Delta t-\frac{1}{2}\overset{\rightarrow}{\mathbf{a}}(\Delta t)^2$ $\overset{\rightarrow}{\mathbf{v}_1}^2=\overset{\rightarrow}{\mathbf{v}_0}^2+2\overset{\rightarrow}{\mathbf{a}} \Delta \overset{\rightarrow}{\mathbf{r}}.$ And also, for acceleration which is not constant, this equation also holds: $\overset{\rightarrow}{\mathbf{v}_1}^2-\overset{\rightarrow}{\mathbf{v}_0}^2=2\int_{t_0}^{t_1} \overset{\rightarrow}{\mathbf{a}} \, d\overset{\rightarrow}{\mathbf{r}}.$ $\large \textbf{\color{blue}{(iii) Free Falling Bodies}}$
A particle which undergoes free fall will experience a acceleration of $-g$ which is the gravitational acceleration, approximately $9.81 \textrm{m/s}^2$ . We can then describe the motion of the particle by using kinematics equations, $a_y=-g,$ $v_y=v_{0y}-gt,$ $y=y_0+v_{0y}t-\frac{1}{2}gt^2.$ $\large \textbf{\color{blue}{(iv) Problems}}$

JEE Mock Test wrote:

Two students simultaneously start from the same place on a circular track and run for $2$ minutes. In this time, one of them completes three and the other four revolutions with uniform speed. Due to thick vegetation in a circular area inside the track, either of the boys can see only one third of the track at a time (a sixth of the track both in front and behind them). How long during their run do they remain visible to each other?
Solution

HRK wrote:

An object, constrained to move along the $x$ axis, travels a distance $d_1$ with constant velocity $v_1$ for a time $t_1$ . It then instantaneously changes its velocity to a constant $v_2$ for a time $t_2$ , travelling a distance $d_2$ .
(a) Show that $\frac{v_1d_1+v_2d_2}{d_1+d_2}\ge \frac{v_1t_1+v_2t_2}{t_1+t_2}.$ (b) Under what condition is this an equality?
Solution

HRK wrote:

A basketball player, about to "dunk" the ball, jumps $76$ cm vertically. How much time does the player spend (a) in the top $15cm$ of this jump and $(b)$ in the bottom $15$ cm?
Solution