Solved Problem on Kinematics
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A body describes on a plane a curve given by the equations:
\[
\begin{gather}
x=3\cos t-\cos 2t\\
y=3\sin t-\sin 2t
\end{gather}
\]
where x and y are given in meters and t in seconds. Determine the speed and
acceleration of the body at instant t = π/2 s.
Solution
The motion of the particle is the vector sum of the displacement along the Ox axis, and the displacement along the Oy axis, the speeds along these axes will be given by
\[ \bbox[#99CCFF,10px]
{v=\frac{dr}{dt}}
\]
\[
v_{x}=\frac{d}{dt}(3\cos t-\cos 2t)
\]
Differentiation of \( 3\cos t-\cos 2t \)
The derivative of a difference is the difference of the derivatives
The derivative of a difference is the difference of the derivatives
\[
\begin{gather}
\frac{d}{dt}(3\cos t-\cos 2t)=\frac{d}{dt}(3\cos t)-\frac{d}{dt}(\cos 2t) \tag{I}
\end{gather}
\]
The derivative of
\( 3\cos t \)
is
\[
\begin{gather}
\frac{d}{dt}(3\cos t)=-3\sin t \tag{II}
\end{gather}
\]
The function
\( \cos 2t \)
is a composite function whose derivative is given by the chain rule
\[
\begin{gather}
\frac{dg[h(t)]}{dt}=\frac{dg}{dh}\;\frac{dh}{dt} \tag{III}
\end{gather}
\]
with
\( g(h)=\cos h \)
and
\( h(t)=2t \),
the derivatives will be
\[
\begin{align}
&\frac{dg}{dh}=-\sin h \tag{IV}\\[5pt]
&\frac{dh}{dt}=2 t^{1-1}=2 \tag{V}
\end{align}
\]
substituting expressions (IV) and (v) into expression (III)
\[
\begin{gather}
\frac{dg[h(t)]}{dt}=-2\sin 2t \tag{VI}
\end{gather}
\]
substituting expressions (II) and (VI) into expression (I)
\[
\frac{d}{dt}(3\cos t-\cos 2t)=-3\sin t-(-2\sin 2t)
\]
\[
\begin{gather}
v_{x}=-3\sin t+2\sin 2t \tag{VII}
\end{gather}
\]
\[
v_{y}=\frac{d}{dt}(3\sin t-\sin 2t)
\]
Differentiation of \( 3\sin t-\sin 2t \)
The derivative of difference is the difference of derivatives
The derivative of difference is the difference of derivatives
\[
\begin{gather}
\frac{d}{dt}(3\sin t-\sin 2t)=\frac{d}{dt}(3\sin t)-\frac{d}{dt}(\sin 2t) \tag{VIII}
\end{gather}
\]
The derivative of
\( 3\sin t \)
is
\[
\begin{gather}
\frac{d}{dt}(3\sin t)=3\cos t \tag{IX}
\end{gather}
\]
The function
\( \sin 2t \)
is a composite function whose derivative is given by the chain rule
\[
\begin{gather}
\frac{dg[h(t)]}{dt}=\frac{dg}{dh}\;\frac{dh}{dt} \tag{X}
\end{gather}
\]
with
\( g(h)=\sin h \)
and
\( h(t)=2t \),
the derivatives will be
\[
\begin{align}
&\frac{dg}{dh}=\cos h \tag{XI}\\[5pt]
&\frac{dh}{dt}=2 \tag{XII}
\end{align}
\]
substituting expressions (XI) and (XII) into expression (X)
\[
\begin{gather}
\frac{dg[h(t)]}{dt}=2\cos 2t \tag{XIII}
\end{gather}
\]
substituting expressions (IX) and (XIII) into expression (VIII)
\[
\frac{d}{dt}(3\sin t-\sin 2t)=3\cos t-2\sin 2t
\]
\[
\begin{gather}
v_{y}=3\cos t-2\cos 2t \tag{XIV}
\end{gather}
\]
for \( t=\dfrac{\pi}{2}\;\text{s} \)
\[
\begin{gather}
v_{x}=-3\sin \frac{\pi}{2}+2\sin 2.\frac{\pi}{2}\\
v_{x}=-3.1+2\sin \pi\\
v_{x}=-3+0\\
v_{x}=-3\;\text{m/s} \tag{XV}
\end{gather}
\]
\[
\begin{gather}
v_{y}=3\cos \frac{\pi}{2}-2\cos 2.\frac{\pi}{2}\\
v_{y}=3.0-2\cos \pi \\
v_{y}=0-2.(-1)\\
v_{y}=2\;\text{
m/s} \tag{XVI}
\end{gather}
\]
from the expressions (XV) and (XVI), the velocity vector will be
\[
\vec{v}={\vec{v}}_{x}+{\vec{v}}_{y}
\]
\[ \bbox[#FFCCCC,10px]
{\vec{v}=-3\;\mathbf{\text{i}}+2\;\mathbf{\text{j}}}
\]
where i and j are the unit vectors in x and y directions. The speed will be
\[
\begin{gather}
v^{2}=v_{x}^{2}+v_{y}^{2}\\
v^{2}=(-3)^{2}+2^{2}\\
v^{2}=9+2\\
v=\sqrt{13\;}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v=3,6\;\text{m/s}}
\]
Accelerations along the axes will be given by
\[ \bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\]
\[
a_{x}=\frac{d}{dt}(-3\sin t+2\sin 2t)
\]
the derivatives of
\( 3\sin t \)
and
\( \sin 2t \)
have already been determined in the expressions (IX) and (XIII) above
\[
\begin{gather}
\frac{dv_{x}}{dt}=-3\cos t+2.2\cos 2t\\
a_{x}=-3\cos t+4\cos 2t \tag{XVII}
\end{gather}
\]
\[
a_{y}=\frac{dv_{y}}{dt}(3\cos t-2\cos 2t)
\]
he derivatives of
\( 3\cos t \)
and
\( \cos 2t \)
have already been determined in expressions (II) and (VI) above
\[
\begin{gather}
\frac{dv_{y}}{dt}=-3\sin t-2.(-2\sin 2t)\\
a_{y}=-3\sin t+4\sin 2t \tag{XVIII}
\end{gather}
\]
for \( t=\dfrac{\pi }{2}\;\text{s} \)
\[
\begin{gather}
a_{x}=-3\cos \frac{\pi}{2}+4\cos 2.\frac{\pi}{2}\\
a_{x}=-3.0+4\cos \pi\\
a_{x}=0+4.(-1)\\
a_{x}=-4\;\text{m/s}^{2} \tag{XIX}
\end{gather}
\]
\[
\begin{gather}
a_{y}=-3\sin \frac{\pi}{2}+4\sin 2.\frac{\pi}{2}\\
a_{y}=-3.1+4\sin \pi\\
a_{y}=-3+4.0\\a_{y}=-3\;\text{m/s} \tag{XX}
\end{gather}
\]
from the expressions (XIX) and (XX), the acceleration vector will be
\[
\vec{a}={\vec{a}}_{x}+{\vec{a}}_{y}
\]
\[ \bbox[#FFCCCC,10px]
{\vec{a}=-4\;\mathbf{\text{i}}-3\;\mathbf{\text{j}}}
\]
The magnitude of the acceleration will be
\[
\begin{gather}
a^{2}=a_{x}^{2}+a_{y}^{2}\\
a^{2}=(-4)^{2}+(-3)^{2}\\
a^{2}=16+9\\
a=\sqrt{25\;}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{a=5\;\text{m/s}^{2}}
\]
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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 .