Solved Problem on One-dimensional Motion
Português
English
advertisement
From two distant points, A and B, depart one car from each point, its motions are described by the following equations, measures in International System of Units (S.I.).
\[
\begin{gather}
S_{A}=10t+\frac{3}{2}t^{2}\\
S_{B}=300-2t^{2}
\end{gather}
\]
a) Determine the distance between cars, when the magnitude of their speeds are equal;b) The speed of each car, when they are at a distance calculated in the previous item.
Problem diagram:
Analyzing the equations given in the problem, we see that cars have acceleration. The equation of the displacement as a function of time with constant acceleration is given by
\[ \bbox[#99CCFF,10px]
{S+S_{0}+v_{0}t+\frac{a}{2}t^{2}}
\]
Fom the equations, we see that the car A the strarts from the origin,
S0A = 0, with initial speed
v0A = 10 m/s if acceleration
aA = 3 m/s2. Car B starts from the rest,
v0B = 0, with 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 car A
\[
\begin{gather}
v_{A}=v_{0 A}+a_{A}t\\
v_{A}=10+3t \tag{I}
\end{gather}
\]
For car 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 will find the instant of time in
which the speeds of the two cars equals, equating expressions (I) and (II)
\[
\begin{gather}
|\;v_{A}\;|=|\;v_{B}\;|\\
|\;10+3t\;|=|\;-4t\;|\\
10+3t=4t\\
4t-3t=10\\
t=10\;\text{s} \tag{III}
\end{gather}
\]
substituting this value in the expressions of the positions given in the problem, we will find the position
in the trajectory in which each car is
For car A
\[
\begin{gather}
S_{A}=10\times 10+\frac{3}{2}10^{2}\\
S_{A}=100+1.5\times 100\\
S_{A}=100+150\\
S_{A}=250\;\text{m}
\end{gather}
\]
For car B
\[
\begin{gather}
S_{B}=300-2\times 10^{2}\\
S_{B}=300-2\times 100\\
S_{B}=300-200\\
S_{B}=100\;\text{m}
\end{gather}
\]
The distance between the two cars will be
\[
\begin{gather}
d=|\;S_{B}-S_{A}|\\
d=|\;100-250\;|\\
d=|\;-150\;|
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{d=150\;\text{m}}
\]
b) Substituting the time found in (III) into expressions (I) and (II)
\[
\begin{gather}
v_{A}=10+3\times 10\\
v_{A}=10+30
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v_{A}=40\;\text{m/s}}
\]
\[
v_{B}=-4\times 10
\]
\[ \bbox[#FFCCCC,10px]
{v_{B}=-40\;\text{m/s}}
\]
Note: We see that in the instant calculated, the velocities are not the same,
they have equal magnitude, but remembering that velocity is a vector, it is described by magnitude
and direction. The magnitudes of the velocities are the same,
\( |\;v_{A}\;|=|\;v_{B}\;|=40\;\text{m/s} \),
but they have different directions, the speed of car A is positive, it moves in the same direction
of the reference frame, the speed of the car B is negative, it moves in the opposite direction of
the referential frame. As two vectors are equal only if they have the same magnitude and direction, these
velocitiess are different.
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .