Solved Problem on One-dimensional Motion
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Two buses depart from distant points, A and B, their motions are described by the following equations
\[
\begin{gather}
S_{A}=3t^{2}\\
S_{B}=300-2t^{2}
\end{gather}
\]
units of the International System of Units (S.I.).a) Determine the distance between the buses when the magnitude of their velocities are equal;
b) The speed of each bus when they are at a distance calculated in the previous item.
Problem diagram:
Analyzing the equations given in the problem, we see that buses have acceleration. The equation of the displacement as a function of time with constant acceleration is given by
\[ \bbox[#99CCFF,10px]
{S+S_{0}+v_{0}t+\frac{a}{2}t^{2}}
\]
From the equations, we see that truck A starts from the origin,
S0A = 0, with initial speed
v0A = 0, and acceleration
aA = 6 m/s2, the truck B starts from the rest,
v0B = 0, from an initial position
S0B = 300 m, and acceleration
aB = −4 m/s2 (Figure 1).
Solution
a) The equation of velocity as a function of time is given by
\[ \bbox[#99CCFF,10px]
{v=v_{0}+a t}
\]
For bus A
\[
\begin{gather}
v_{A}=v_{0 A}+a_{A}t\\
v_{A}=6t \tag{I}
\end{gather}
\]
For bus B
\[
\begin{gather}
v_{B}=v_{0 B}+a_{B}t\\
v_{B}=0-4t\\
v_{B}=-4t \tag{II}
\end{gather}
\]
Imposing the condition that the magnitude of velocities must be equal, we find the instant o time in which
the speeds of the two buses equals, equating expressions (I) and (II)
\[
\begin{gather}
|\;v_{A}\;|=|\;v_{B}\;|\\
|\;6t\;|=|\;-4t\;|\\
6t=4t\\6t-4t=0\\
2t=0\\t=\frac{0}{2}\\
t=0 \tag{III}
\end{gather}
\]
substituting this value in the expressions of the displacement given in the problem, we will find the
position in which each of the buses is
For bus A
\[
\begin{gather}
S_{A}=3\times 0^{2}\\
S_{A}=3\times 0\\
S_{A}=0
\end{gather}
\]
For bus B
\[
\begin{gather}
S_{B}=300-2\times 0^{2}\\
S_{B}=300-2\times 0\\
S_{B}=300-0\\
S_{B}=300\;\text{m}
\end{gather}
\]
The distance between the two buses will be
\[
\begin{gather}
d=|\;S_{B}-S_{A}|\\
d=|\;300-0\;|\\
d=|\;300\;|
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{d=300\;\text{m}}
\]
b) Substituting the time found in (III) into expressions (I) and (II)
\[
v_{A}=6\times 0
\]
\[ \bbox[#FFCCCC,10px]
{v_{A}=0}
\]
\[
v_{B}=-4\times 0
\]
\[ \bbox[#FFCCCC,10px]
{v_{B}=0}
\]
Note: As the two buses come from the origin and have different accelerations, the only
instant of time in which their speeds are equal happens when both are stopped and their distance is equal
to the distance between their initial positions.
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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 .