Solved Problem on One-dimensional Motion
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A particle travels along a path in an accelerated motion. At the instant t = 2 s its velocity is equal to −4 m/s, at the instant t = 3 s its position is 4.5 m, and at the instants t = 5 s and t = 7 s its velocities are equal, and of opposite signs. Determine the function S=f(t) of the motion.
Problem data:
- Velocity at t = 2 s: v(2) = −4 m/s;
- Position at time t = 3 s: S(3) = 4.5 m;
- Velocity at t = 5 s: v(5) = v;
- Velocity at t = 7 s: v(7) = −v.
The problem gives us the following conditions:
- v(2) = −4 m/s (I)
- S(3) = 4.5 m (II)
- v(5) = v (III)
- v(7) = −v (IV)
The function of displacement as a function of time with constant acceleration is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S(t)=S_{0}+v_{0}t+\frac{a}{2}t^{2}} \tag{V}
\end{gather}
\]
The equation of velocity as a function of time is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v(t)=v_{0}+a} \tag{VI}
\end{gather}
\]
Substituting conditions (I), (III), and (IV) into equation (VI)
\[
\begin{gather}
v(2)=-4=v_{0}+2a \tag{VII}
\end{gather}
\]
\[
\begin{gather}
v(5)=v=v_{0}+5a \tag{VIII}
\end{gather}
\]
\[
\begin{gather}
v(7)=-v=v_{0}+7a \tag{IX}
\end{gather}
\]
in equations (VIII) and (IX), v and −v are velocities of the same magnitude and opposite
signs at instants 5 and 7 seconds. Equations (VII), (VIII), and (IX) can be written as a system of three
equations with three unknowns (v0, a and v).
\[
\begin{gather}
\left\{
\begin{matrix}
v_{0}+2a=-4\\
v_{0}+5a=v\\
v_{0}+7a=-v
\end{matrix}
\right.
\end{gather}
\]
solving the first equation of the system for v0
\[
\begin{gather}
v_{0}=-4-2a \tag{X}
\end{gather}
\]
adding the second and third equations of the system
\[
\begin{align}
v_{0}+5a &=v\\
(\text{+}) \qquad v_{0}+7a &=-v\\
\hline
2v_{0}+12a &=0 \tag{XI}
\end{align}
\]
substituting equation (X) in equation (XI)
\[
\begin{gather}
2(-4-2a)+12a=0\\[5pt]
-8-4a+12a=0\\[5pt]
8a=8\\[5pt]
a=\frac{8}{8}\\[5pt]
a=1\;\text{m/s}^{2} \tag{XII}
\end{gather}
\]
substituting the result (XII) for the acceleration into equation (X)
\[
\begin{gather}
v_{0}=-4-2.1\\[5pt]
v_{0}=-4-2\\[5pt]
v_{0}=-6\;\text{m/s} \tag{XIII}
\end{gather}
\]
Substituting condition (II) into equation (V)
\[
\begin{gather}
S(3)=4.5=S_{0}+3\times(-6)+\frac{1}{2}\times3^{2}\\[5pt]
S_{0}-18+\frac{1}{2}\times9=4.5\\[5pt]
S_{0}-18+4.5=4.5\\[5pt]
S_{0}=18-4.5+4.5\\[5pt]
S_{0}=18\;\text{m} \tag{XIV}
\end{gather}
\]
substituting the results (XII), (XIII), and (XIV) in equation (V)
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{S(t)=18-6t+\frac{1}{2}t^{2}}
\end{gather}
\]
Note: We did not calculate the magnitude of the velocity v that appears in conditions
(III) and (IV) because it is not part of the function S(t). Out of curiosity, we can
calculate the value of v by substituting the results (XII) and (XIII) in the second equation of the
system
\[
\begin{gather}
v=-6+5.1\\[5pt]
v=-6+5\\[5pt]
v=-1\;\text{m/s}
\end{gather}
\]
The velocities at instants 5 and 7 seconds will be
\[
\begin{align}
& v(5)=v=-1\text{m/s}\\[10pt]
& v(7)=-v=-(-1)=1\;\text{m/s}
\end{align}
\]
the signs of the velocities in conditions (III) and (IV) were randomly assigned, the true signs are
opposite to the chosen ones.
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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 .