Motion of the particle
The particle motion relative to the fixed frame
Consider the particle A in space which is defined at any time \(t\) by the three Cartesian coordinates \(x(t)\),\(y(t)\), and \(z(t)\) as shown in Figure 1. In order to describe the motion of the particle A in three dimensional space the position vector must be derived which
can be written in the following form $$ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}.$$
Figure 1 - The motion of the particle A on the curve C.
Velocity of the particle
The velocity of the particle A in three dimensional space is defined as the time rate of change of the position. The velocity vector of the particle A can be written in the following form $$ \mathbf{v}(t) = \frac{\mathrm{d}\mathbf{r}(t)}{\mathrm{d}t}
= \dot{x}(t) + \dot{y}(t) + \dot{z}(t) = v_x(t)\mathbf{i} + v_y(t)\mathbf{j} + v_z(t)\mathbf{k}$$ where: $$ v_x(t) = \dot{x}(t), \quad v_y(t) = \dot{y}(t), \quad v_z(t) = \dot{z}(t),$$ represent the Cartesian components of the velocity vector.
Acceleration of the particle
The acceleration of the particle A in three dimensional space is defined as the time rate of change of the velocity. The acceleration vector of the particle A can be written in the following form $$ \mathbf{a}(t) = \frac{\mathrm{d}\mathbf{v}(t)}{\mathrm{d}t}
= \dot{v}_x(t)\mathbf{i} + \dot{v}_y(t)\mathbf{j} + \dot{v}_z(t)\mathbf{k}, $$ where $$ a_x(t) = \dot{v}_x(t) = \ddot{x}(t), \quad a_y(t) = \dot{v}_y(t) = \ddot{y}(t), \quad a_z(t) = \dot{v}_z(t) = \ddot{z}(t),$$ represents the Cartesian components
of the acceleration vector.
Rectilinear motion
The rectilinear motion is the motion along a straight line. In rectilinear motion the three-dimensional equatiosn of position, speed, and acceleration are reduced to only one dimension so the vector notation can be replaced by scalar components. Consider
the distance of the particle A from the fixed origin with \(s(t)\). Then the velocity of the particle A can be written in the following form $$ v(t) = \frac{\mathrm{d}s(t)}{\mathrm{d}t} = \dot{s}(t).$$ The acceleration of particle A can be
written as $$ a(t) = \frac{\mathrm{d}v(t)}{\mathrm{d}t} = \frac{\mathrm{d}^2s(t)}{\mathrm{d}t^2} = \ddot{s}(t).$$
So far, in derived expressions, the distance, the velocity, and the acceleration are explicit function of \(t\). To derive a relation among s,v, and a in wich the time t is only implicit using the chain rule. With application of the chain rule for differentiation the following expression is obtained: $$ a = \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = \frac{dv}{ds}v$$ or $$a\mathrm{d}s = v\mathrm{d}v.$$ Integrating previous equation between \(s=s_1, v=v_1\), and \(s=s_2,v=v_2\) the following expression is obtained $$\int_{s_1}^{s_2} a\mathrm{d}s = \int_{v_1}^{v_2} = \frac{1}{2} \left(v_2^2 - v_1^2\right). $$
So far, in derived expressions, the distance, the velocity, and the acceleration are explicit function of \(t\). To derive a relation among s,v, and a in wich the time t is only implicit using the chain rule. With application of the chain rule for differentiation the following expression is obtained: $$ a = \frac{dv}{dt} = \frac{dv}{ds}\frac{ds}{dt} = \frac{dv}{ds}v$$ or $$a\mathrm{d}s = v\mathrm{d}v.$$ Integrating previous equation between \(s=s_1, v=v_1\), and \(s=s_2,v=v_2\) the following expression is obtained $$\int_{s_1}^{s_2} a\mathrm{d}s = \int_{v_1}^{v_2} = \frac{1}{2} \left(v_2^2 - v_1^2\right). $$
Uniform motion
In the case of the unifom motion the speed of the particle A is constant (\(v(t) = v_0 = \mathrm{const.}\)). By integration, the following expression is obtained $$ s(t) = \int v\mathrm{v}t = v_0 t + C_1.$$ With the initial condition \(s(t=t_1) = s_1\)
the \(C_1 = s_1 - v_0 t_1\) so the equation in the final form $$ s(t) = v_0(t-t_1) + s_1.$$
Uniform accelerated motion
In the case of the uniform accelerated motion the particle A is moving with constant acceleration \(a(t) = a_0 = \mathrm{const.}\). By integration the following expression is obtained $$ v(t) = \int a\mathrm{d}t = a_0t + C_1$$ or $$ s(t) = \int v\mathrm{d}t
= a_0\frac{t^2}{2} + C_1t + C_2.$$
The initial conditions in case of uniform accelerated motion are $$ v(t=t_1) = v_1, s(t=t_1) = s_1.$$ By using these initial condition the constats \(C_1\) and \(C_2\) are equal to $$ C_1 = v_1 - a_0t_1, C_2 = s_1 - v_1t_1 + \frac{a_0t_1^2}{2}.$$ Substituting these constants into the equation the following expression is obtained $$ s(t) = a_0 \frac{(t-t_1)^2}{2} + v_1(t-t_1) + s_1.$$
The initial conditions in case of uniform accelerated motion are $$ v(t=t_1) = v_1, s(t=t_1) = s_1.$$ By using these initial condition the constats \(C_1\) and \(C_2\) are equal to $$ C_1 = v_1 - a_0t_1, C_2 = s_1 - v_1t_1 + \frac{a_0t_1^2}{2}.$$ Substituting these constants into the equation the following expression is obtained $$ s(t) = a_0 \frac{(t-t_1)^2}{2} + v_1(t-t_1) + s_1.$$
Nonuniform Accelerated Motion
In this case, the acceleration is equal to \(a_0 = f_0(t)\). By integration the equations for speed and position are obtained $$ v(t) = \int a(t) \mathrm{d}t = \int f_0\mathrm{d}t = f_1(t) + C_1$$ or $$ s(t) = \int v(t) \mathrm{d}t = \int \left(f_1(t)
+ C_1\right)\mathrm{d}t = f_2(t) + C_1t + C_2.$$ The constants \(C_1\), and \(C_2\) are determined from the initial conditions or equivalent conditions.
Problems
- A particle is moving on a straight line with the acceleration of \(a = (2t-6) \left[\frac{\mathrm{m}}{\mathrm{s}^2}\right]\). The particle is start moving from rest. Determine the particle's velocity when \(t = 6 \left[s\right]\) and what is its position when \(t = 11 \left[s\right]\) ? Solution: The particle is starting from rest which means that at \(t = 0 \left[s\right]\) the \(v = 0 \left[\frac{m}{s}\right]\) and \(a = 0\). The particle velocity at t = 6 s ? $$ a = \frac{dv}{dt} = 2t - 6 $$ $$ \int_{0}^{v} dv = \int_{0}^{t}(2t - 6) $$ $$ v = 2 \frac{t^2}{2} - 6t$$ $$ v = t^2 - 6t $$ $$ v = 36 - 36$$ $$ v = 0 $$ Solution: The position of the particle at \(t = 11 \left[s\right]\) ? $$ v = \frac{ds}{dt} = t^2 - 6t$$ $$ s = \int_{0}^{t} (t^2 - 6t) dt$$ $$ s = \frac{t^3}{3} - 6\frac{t^2}{2} $$ $$ s = \frac{t^3}{3} - 3t^2 $$ $$ s = 443.67 - 363 = 70.7 [m]$$
- The initial velocity of the particle is equal to 12 m/s to the right. At the s_0 = 0, determine its position when t= 10 s if a = 2 m/s^2 to the left. $$ \begin{eqnarray}s &=& s_0 + v_0t - \frac{at^2}{2}\\ \nonumber s &=& 0 + 12*10 - \frac{2*10^2}{2} \\ \nonumber s &=& 120 - 100 \\ \nonumber s &=& 20 [m]\end{eqnarray}$$
- A particle travles along a straight line with a velocity \(v = (12-3t^2) \left[\frac{\mathrm{m}}{\mathrm{s}}\right]\) When \(t = 1 [s]\) the particle is located 10 [m] to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to 10 s. $$\begin{eqnarray}v &=& (12-3t^2)\\\nonumber \mathrm{for}\quad t &=& 4 \left[s\right] => a = ?\\\nonumber \frac{\mathrm{d}v}{\mathrm{d}t} &=& 0 - 3\cdot 2 \cdot t \\\nonumber a_{t=4} &=& -6\cdot 4 = -24 \left[\frac{\mathrm{m}}{\mathrm{s}^2}\right]\end{eqnarray}$$
- The velocity of the particle traveling in a straight line is given by v = (6t - 3t^2). The s = 0 when t = 0 determine the particle's deceleration and position when t = 3s. How far is the particle traveled during the 3s time interval, and what is the average speed ? $$v = \frac{ds}{dt}$$ $$ s = 3t^2 - t^3$$ $$ s_{t=3} = 27-27 = 0$$ $$ a = \frac{dv}{dt} = 6-6t $$ $$ a = 6-18 = -12 $$ $$ $$
At \(t = 0 \left[s\right],\) \(x_{t=0}=?\) and at \(t=10 \left[s\right],\) \(x_{t=10}=?\) $$\begin{eqnarray} v &=&\frac{\mathrm{d}x}{\mathrm{d}t} = 12 - 3 t^2/\int\\\nonumber \int \mathrm{d}x &=& \int \left(12 - 3\cdot t^2\right) \mathrm{d}t \\\nonumber x &=& 12 t - t^3 \\\nonumber x_{t=0} &=& 0 \left[m\right] \\\nonumber x_{t=10} &=& 12\cdot 10 - 1000 = 120 - 1000 = -880 \left[m\right]\\\nonumber \Delta x &=& x_{t=10} - x_{t=0}\\\nonumber \Delta x &=& 880 - 0 = 880 \left[m\right] \end{eqnarray} $$
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