Solved Problem on Kinematics
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Particle B starts a motion from point O at the same time particle A passes through
this point. Both particles run the same straight path and the curves in the velocity-time graph are a
quarter of circumference with equal radius, as shown in the figure. Determine:
a) The time in which particles have the same speed;
b) The value of this speed;
c) The time in particles have the same acceleration in magnitude;
d) The value of this acceleration.
a) The time in which particles have the same speed;
b) The value of this speed;
c) The time in particles have the same acceleration in magnitude;
d) The value of this acceleration.
Solution
The equation of a circle is given by
\[ \bbox[#99CCFF,10px]
{(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2}}
\]
where x0 and y0 are the coordinates of the center of the circle and
r is the radius.As the graph of the velocities as a function of time are circular arcs, we can make the following associations x = t, y = v and r = 10.
For particle B, we have a circle centered on origin
(x0, y0) = (0, 0) (Figure 1), so the speed equation will be
\[
\begin{gather}
t^{2}+v_{B}^{2}=10^{2}\\
v_{B}^{2}=100-t^{2} \tag{I}\\
v_{B}=\sqrt{100-t^{2}\;} \tag{I}
\end{gather}
\]
Figure 1
For particle A, we have a circle centered on
(x0, y0) = (10, 0) (Figure 2), so the speed equation will be
\[
\begin{gather}
(t-10)^{2}+v_{A}^{2}=10^{2}\\
v_{A}^{2}=100-(t-10)^{2} \tag{III}\\
v_{A}=\sqrt{100-(t-10)^{2}\;} \tag{IV}
\end{gather}
\]
Figure 2
a) Under the condition that speeds are equal and using expressions (I) and (III)
\[
\begin{gather}
v_{B}^{2}=v_{A}^{2}\\
100-t^{2}=100-(t-10)^{2}\\
t^{2}=(t-10)^{2}\\
t^{2}=t^{2}-20t+100\\
-20t-100=0\\
t=\frac{100}{20}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{t=5\;\text{s}}
\]
b) Substituting the result of the previous item in the expression (II)
\[
\begin{gather}
v_{B}=\sqrt{100-5^{2}\;}\\
v_{B}=\sqrt{75\;}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v_{A}=v_{B}=8.7\;\text{m/s}}
\]
c) The acceleration is given by
\[ \bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\]
differentiating expressions (II) and (IV) with repect to time, we have the accelerations aA
and aB of the particles.
Differentiation of \( v_{B}=\sqrt{100-t^{2}\;} \)
the function vB(t) is a composite function whose derivative is given by the chain rule
the function vB(t) is a composite function whose derivative is given by the chain rule
\[
\begin{gather}
\frac{dv[u(t)]}{dt}=\frac{dv}{du}\frac{du}{dt} \tag{V}
\end{gather}
\]
with
\( v(u)=\sqrt{u\;} \)
and
\( u(t)=100-t^{2} \),
so the derivatives will be
\[
\begin{gather}
\frac{dv}{du}=u^{\frac{1}{2}}=\frac{1}{2}u^{\frac{1}{2}-1}=\frac{1}{2}u^{-{\frac{1}{2}}}=\frac{1}{2\sqrt{u\;}} \tag{VI}
\end{gather}
\]
\[
\begin{gather}
\frac{du}{dt}=-2t \tag{VII}
\end{gather}
\]
substituting expressions (VI) and (VII) into expression (V)
\[
\frac{dv_{B}}{dt}=\frac{1}{2\sqrt{u\;}}(-2t)=\frac{-{t}}{\sqrt{100-t^{2}\;}}
\]
\[
\begin{gather}
a_{B}=-{\frac{t}{\sqrt{100-t^{2}\;}}} \tag{VIII}
\end{gather}
\]
Differentiation of \( v_{A}=\sqrt{100-(t-10)^{2}\;} \)
the function vA(t) is a composite function whose derivative is given by the chain rule
the function vA(t) is a composite function whose derivative is given by the chain rule
\[
\begin{gather}
\frac{dv[u(t)]}{dt}=\frac{dv}{du}\frac{du}{dt} \tag{IX}
\end{gather}
\]
with
\( v(u)=\sqrt{u\;} \)
and
\( u(t)=100-(t-10)^{2} \),
so the derivatives will be
\[
\begin{gather}
\frac{dv}{du}=u^{\frac{1}{2}}=\frac{1}{2}u^{\frac{1}{2}-1}=\frac{1}{2}u^{-{\frac{1}{2}}}=\frac{1}{2\sqrt{u\;}} \tag{X}
\end{gather}
\]
\[
\begin{gather}
\frac{du}{dt}=-2(t-10) \tag{XI}
\end{gather}
\]
substituting expressions (X) and (XI) into expression (IX)
\[
\frac{dv_{A}}{dt}=\frac{1}{2\sqrt{u\;}}[-2(t-10)]=\frac{-(t-10)}{\sqrt{100-(t-10)^{2}}}
\]
\[
\begin{gather}
a_{A}=-{\frac{t-10}{\sqrt{100-(t-10)^{2}}}} \tag{XII}
\end{gather}
\]
Under the condition that accelerations are equal and using expressions (VIII) and (XII)
\[
\begin{gather}
a_{B}=a_{A}\\[5pt]
-{\frac{t}{\sqrt{100-t^{2}}}}=-{\frac{t-10}{\sqrt{100-(t-10)^{2}}}}\\[5pt]
\left[-{\frac{t}{\sqrt{100-t^{2}}}}\right]^{2}=\left[-{\frac{t-10}{\sqrt{100-(t-10)^{2}}}}\right]^{2}\\[5pt]
\frac{t^{2}}{100-t^{2}}=\frac{(t-10)^{2}}{100-(t-10\;)^{2}\;}\\[5pt]
t^{2}\left[100-(t-10\;)^{2}\right]=(t-10)^{2}(100-t^{2})\\[5pt]
100t^{2}-\cancel{[t^{2}(t-10)^{2}]}=100(t-10)^{2}-\cancel{[t^{2}(t-10)^{2}]}\\[5pt]
100t^{2}=100(t-10)^{2}\\[5pt]
t^{2}=t^{2}-20t+100\\[5pt]
-20t+100=0\\[5pt]
t=\frac{100}{20}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{t=5\;\text{s}}
\]
d) Substituting the result of the previous item in the expression (VIII)
\[
\begin{gather}
a_{B}=-\frac{{5}}{\sqrt{100-5^{2}\;}}\\
a_{B}=-\frac{{5}}{\sqrt{100-25\;}}\\
a_{B}=-\frac{{5}}{\sqrt{75\;}}\\
a_{B}=-\frac{{5}}{8.7}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{a_{B}\simeq -0.6\;\text{m/s}^{2}}
\]
Substituting the result of the previous item in the expression (XIII)
\[
\begin{gather}
a_{A}=-\frac{{5-10}}{\sqrt{100-(5-10)^{2}\;}}\\
a_{A}=-\frac{{-5}}{\sqrt{100-(-5)^{2}\;}}\\
a_{A}=\frac{5}{\sqrt{100-25\;}}\\
a_{A}=\frac{5}{\sqrt{75\;}}\\
a_{B}=\frac{-{5}}{8.7}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{a_{A}\simeq 0.6\;\text{m/s}^{2}}
\]
Note: We see that the accelerations of the particles are equal in magnitude
(0,6 m/s2), but they have opposite signs, while particle B is reducing its speed
(braking), particle A is increasing the speed (accelerating), which agrees with the curves shown
on the problem graph.
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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 .