Solved Problem on Kinematics
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A particle describes a curve in a plane, such as its positions relative to an orthogonal Cartesian system, taken in that plan, vary with time according to equations:
\[
\begin{gather}
x=t^{3}-2t\\[8pt]
y=4t^{2}
\end{gather}
\]
where x and y are in meters and t in seconds. Determine the speed and acceleration of
the point at instant t = 2 s.
Solution
The motion of the particle is the vector sum of motion along the Ox axis and another motion along the Oy axis, the speeds along these axes will be given by
\[ \bbox[#99CCFF,10px]
{v=\frac{dr}{dt}}
\]
\[
\begin{gather}
\frac{dx}{dt}=3t^{3-1}-2^{2-1}\\
v_{x}=3t^{2}-2 \tag{I}
\end{gather}
\]
\[
\begin{gather}
\frac{dy}{dt}=2\times 4t^{2-1}\\
v_{y}=8t \tag{II}
\end{gather}
\]
for t = 2 s
\[
\begin{gather}
v_{x}=3\times 2^{2}-2\\
v_{x}=3\times 4-2\\
v_{x}=12-2\\
v_{x}=10\;\text{m/s} \tag{III}
\end{gather}
\]
\[
\begin{gather}
v_{y}=8\times 2\\
v_{y}=16\;\text{m/s} \tag{IV}
\end{gather}
\]
with expressions (III) and (IV), the vector velocity will be
\[
\vec{v}={\vec{v}}_{x}+{\vec{v}}_{y}
\]
\[ \bbox[#FFCCCC,10px]
{\vec{v}=10\;\mathbf{\text{i}}+16\;\mathbf{\text{j}}}
\]
Where i and j are the unit vectors in x and y directions. The speed will be
\[
\begin{gather}
v^{2}=v_{x}^{2}+v_{y}^{2}\\
v^{2}=10^{2}+16^{2}\\
v^{2}=100+256\\
v=\sqrt{356\;}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v=18.9\;\text{m/s}}
\]
The accelerations along the axes will be given by
\[ \bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\]
With expressions (I) and (II), we obtain the accelerations in the directions i and j
\[
\begin{gather}
\frac{dv_{x}}{dt}=2\times 3t^{2-1}-0\\
a_{x}=6t \tag{V}
\end{gather}
\]
\[
\begin{gather}
\frac{dv_{y}}{dt}=8t^{1-1}\\
a_{y}=8 \tag{VI}
\end{gather}
\]
for t = 2 s
\[
\begin{gather}
a_{x}=6\times 2\\
a_{x}=12\;\text{m/s}^{2} \tag{VII}
\end{gather}
\]
\[
\begin{gather}
a_{y}=8\;\text{m/s}^{2} \tag{VIII}
\end{gather}
\]
with expressions (VII) and (VIII), the acceleration vector will be
\[
\vec{a}={\vec{a}}_{x}+{\vec{a}}_{y}
\]
\[ \bbox[#FFCCCC,10px]
{\vec{a}=12\;\mathbf{\text{i}}+8\;\mathbf{\text{j}}}
\]
The magnitude of the acceleration will be
\[
\begin{gather}
a^{2}=a_{x}^{2}+a_{y}^{2}\\
a^{2}=12^{2}+8^{2}\\
a^{2}=144+64\\
a=\sqrt{208\;}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{a=14.4\;\text{m/s}^{2}}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .