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.


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
\[ \begin{gather} \frac{dv[u(t)]}{dt}=\frac{dv}{du}\frac{du}{dt} \tag{V} \end{gather} \]
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
\[ \begin{gather} \frac{dv[u(t)]}{dt}=\frac{dv}{du}\frac{du}{dt} \tag{IX} \end{gather} \]
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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