Solved Problem on One-dimensional Motion
ES
FR
PT-BR
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Two particles move along the same line, with motion given by the following equations
\[
\begin{gather}
S_1=-10t+5t^2\\[10pt]
S_2=30+5t-10t^2
\end{gather}
\]
the positions are given in centimeters from the origin, and time t is given in seconds. Determine:a) The instant of time when the two particles meet;
b) The velocities and accelerations of both at that instant;
c) The position of the meeting point;
d) When and where the velocities of the two particles are equal;
e) The instant of time when the particles change direction.
Solution
The equations given in the problem are equations of displacement with constant acceleration
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_0+v_0 t+\frac{a}{2}t^2}
\end{gather}
\]
Making the following associations
\[
\begin{array}{c}
S_1 & = & 0 & - & 10 & t & + & 5 & t^2\\
\downarrow & & \downarrow & & \downarrow & & & \downarrow \\
S & = & S_0 & + & v_0 & t & + & \dfrac{a}{2} & t^2
\end{array}
\]
the initial position of particle 1 is S01 = 0, the initial velocity
v01 = −10 cm/s and the acceleration
\( \frac{a_{1}}{2}=5\Rightarrow a_{1}=2\times 5\Rightarrow a_{1}=10\;\text{cm/s}^{2} \)
\[
\begin{gather}
S_2 & = & 30 & + & 5 & t & - & 10 & t^2\\
\downarrow & & \downarrow & & \downarrow & & & \downarrow & \\
\;S & = & S_0 & + & v_0 & t & + & \dfrac{a}{2} & t^2
\end{gather}
\]
the initial position of particle 2 is be S02 = 30 cm, the initial velocity
v02 = 5 cm/s and the acceleration
\( \frac{a_{2}}{2}=-10\Rightarrow a_{2}=-2\times 10\Rightarrow a_{2}=-20\;\text{cm/s}^{2} \)
a) At the instant the particles meet, they are in the same position, applying the condition that the position of the two particles are equals
\[
\begin{gather}
S_1=S_2\\[5pt]
-\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)t+\left(5\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2=30\;\mathrm{cm}+\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)t-\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2\\[5pt]
-\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)t+\left(5\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2-30\;\mathrm{cm}-\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)t+\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2=0\\[5pt]
\left(15\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2+\left(15\;\mathrm{\small{\frac{cm}{s}}}\right)t-30\;\mathrm{cm}=0
\end{gather}
\]
dividing the equation by 15
\[
\begin{gather}
\qquad\qquad \left(15\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2+\left(15\;\mathrm{\small{\frac{cm}{s}}}\right)t-30\;\mathrm{cm}=0 \qquad (\div15)\\[5pt]
\left(1\;\mathrm{\small{\frac{cm}{s^2}}}\right)t^2+\left(1\;\mathrm{\small{\frac{cm}{s}}}\right)t-2\;\mathrm{cm}=0
\end{gather}
\]
This is a Quadratic Equation where the unknown is the desired value of time.
Solution of the Quadratic Equation
\( t^{2}-t-2=0 \)
\[
\begin{gather}
\Delta=b^2-4ac=(-1)^2-4\times 1\times(-2)=1+8=9\\[5pt]
t=\frac{-b\pm\sqrt{\Delta \;}}{2a}=\frac{-(-1)\pm \sqrt{9\;}}{2\times 1}=\frac{1\pm 3}{2}
\end{gather}
\]
the two roots of the equation are
\[
\begin{gather}
t_1=2\;\mathrm s\\
e\\
t_2=-1\;\mathrm s
\end{gather}
\]
As there is no negative time, we neglect the second root, and the meeting instant will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t=2\;\mathrm s}
\end{gather}
\]
b) The equation of velocity as a function of time is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=v_0+at}
\end{gather}
\]
- Particle 1:
\[
\begin{gather}
v_1=v_{01}+a_1t\\[5pt]
v_1=-\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)+\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)t \tag{I}
\end{gather}
\]
substituting the instant of time found in part (a)
\[
\begin{gather}
v_1=-\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)+\left(10\;\mathrm{\small{\frac{cm}{s^\cancel 2}}}\right)(2\;\mathrm{\cancel s})
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_1=10\;\mathrm{cm/s}}
\end{gather}
\]
The acceleration is constant and equal to the
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_1=10\;\mathrm{cm/s^2}}
\end{gather}
\]
- Particle 2:
\[
\begin{gather}
v_2=v_{02}+a_2t\\[5pt]
v_2=\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)-\left(20\;\mathrm{\small{\frac{cm}{s^2}}}\right)t\tag{II}
\end{gather}
\]
substituting the instant of time found in part (a)
\[
\begin{gather}
v_2=\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)-\left(20\;\mathrm{\small{\frac{cm}{s^\cancel 2}}}\right)(2\;\mathrm{\cancel s})
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_2=-35\;\mathrm{cm/s}}
\end{gather}
\]
The acceleration is constant and equal to the
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_2=-20\;\mathrm{cm/s^2}}
\end{gather}
\]
c) Substituting the instant of meeting in the equation of particle 1
\[
\begin{gather}
S_1=-\left(10\;\mathrm{\small{\frac{cm}{\cancel s}}}\right)(2\;\mathrm{\cancel s})+\left(5\;\mathrm{\small{\frac{cm}{s^2}}}\right)(2\;\mathrm s)^2\\[5pt]
S_1=-20\;\mathrm{cm}+\left(5\;\mathrm{\small{\frac{cm}{\cancel{s^2}}}}\right)(4\;\mathrm{\cancel{s^2)}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S_1=0}
\end{gather}
\]
the particles meet at the origin of the system.
Note: if we had substituted in the equation of particle 2, the result would be the same
\[
\begin{gather}
S_2=30\;\mathrm{cm}+\left(5\;\mathrm{\small{\frac{cm}{\cancel s}}}\right)(2\;\mathrm{\cancel s})-\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)(2\;\mathrm s)^2\\[5pt]
S_2=30\;\mathrm{cm}+10\;\mathrm{cm}-\left(10\;\mathrm{\small{\frac{cm}{\cancel{s^2}}}}\right)(4\mathrm{\cancel{cm^2}})\\[5pt]
S_2=0
\end{gather}
\]
d) Equating equations (I) and (II), we have the instant when the particle velocities are equal
\[
\begin{gather}
v_{1}=v_{2}\\[5pt]
-\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)+\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)t=\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)-\left(20\;\mathrm{\small{\frac{cm}{s^2}}}\right)t\\[5pt]
\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)t+\left(20\;\mathrm{\small{\frac{cm}{s^2}}}\right)t=\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)+\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)\\[5pt]
t=\frac{15\;\mathrm{\frac{\cancel{cm}}{\cancel s}}}{30\;\mathrm{\frac{\cancel{cm}}{s^\cancel 2}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t=0.5\;\mathrm s}
\end{gather}
\]
Substituting this instant of time in the equations given in the problem
\[
\begin{gather}
S_1=-\left(10\;\mathrm{\small{\frac{cm}{\cancel s}}}\right)(0.5\;\mathrm{\cancel s})+\left(5\;\mathrm{\small{\frac{cm}{s^2}}}\right)(0.5\;\mathrm s)^{2}\\[5pt]
S_1=-5\;\mathrm{cm}+\left(5\;\mathrm{\small{\frac{cm}{\cancel{s^2}}}}\right)(0.25\;\mathrm{\cancel{s^2}})
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S_1=-3.75\;\mathrm{cm}}
\end{gather}
\]
\[
\begin{gather}
S_2=30\;\mathrm{cm}+\left(5\;\mathrm{\small{\frac{cm}{\cancel s}}}\right)(0.5\;\mathrm{\cancel s})- \left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)(0.5\;\mathrm s)^{2}\\[5pt]
S_2=30\;\mathrm{cm}+2.5\;\mathrm{cm}-\left(10\;\mathrm{\small{\frac{cm}{\cancel{s^2}}}}\right)(0.5\;\mathrm{\cancel{s^2}})
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S_2=30\;\mathrm{cm}}
\end{gather}
\]
e) At the instant the particles change direction, their velocities are equal to zero (v1 = 0 and v2 = 0), substituting these conditions in equations (I) and (II)
\[
\begin{gather}
0=-\left(10\;\mathrm{\small{\frac{cm}{s}}}\right)+\left(10\;\mathrm{\small{\frac{cm}{s^2}}}\right)t\\[5pt]
t=\frac{10\;\mathrm{\frac{\cancel{cm}}{\cancel s}}}{10\;\mathrm{\frac{\cancel{cm}}{s^\cancel 2}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t=1\;\text{s}}
\end{gather}
\]
\[
\begin{gather}
0=\left(5\;\mathrm{\small{\frac{cm}{s}}}\right)-\left(20\;\mathrm{\small{\frac{cm}{s^2}}}\right)t\\[5pt]
t=\frac{5\;\mathrm{\frac{\cancel{cm}}{\cancel s}}}{20\;\mathrm{\frac{\cancel{cm}}{s^\cancel 2}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t=0.25\;\mathrm s}
\end{gather}
\]
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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 .