Solved Problem on One-dimensional Motion
Português
English
advertisement
A train leaves station A, where it is at rest, with a constant acceleration equal to a, at one point, the train driver starts to brake the train with a deceleration equal to b, the train stops at station B. The distance between the stations is equal to L. Determine the total time it takes to travel from one station to another.
Problem data:
- Acceleration of the train: αA = a;
- Deceleration of the train: αB = b;
- Distance between stations: L.
We choose a frame of reference with origin at station A, position D is where the conductor begins to brake the train, and L is the position where station B is located (Figure 1). The quantities with index A refer to the first part of the route and with index B to the second part.
Solution
The train has an acceleration, a > 0 in the first part of the motion and a < 0 in the second part. In each part of the motion, the acceleration is constant, the equation for displacement is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_{0}+v_{0}t+\frac{\alpha }{2}t^{2}} \tag{I}
\end{gather}
\]
for the speed
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=v_{0}+\alpha t} \tag{II}
\end{gather}
\]
We have S0A = 0, which is the position where the train starts the motion and begins
to accelerate. In SA = D is the final position where the train stops to accelerate
and starts to slow down. The train starts from the rest, v0A = 0, with acceleration
αA = a and tA the interval of time at which the train remains in
accelerated motion, substituting this data in the expression (I)
\[
\begin{gather}
S_{A}=S_{0 A}+v_{0 A}t_{A}+\frac{\alpha_{A}}{2}t_{A}^{2}\\[5pt]
D=0+0t_{A}+\frac{a}{2}t_{A}^{2}\\[5pt]
D=\frac{a}{2}t_{A}^{2} \tag{III}
\end{gather}
\]
For the speed, we have that vA is the final speed of the train after accelerating for a
time tA
\[
\begin{gather}
v_{A}=v_{0 A}+\alpha_{A}t_{A}\\[5pt]
v_{A}=0+at_{A}\\[5pt]
v_{A}=at_{A} \tag{IV}
\end{gather}
\]
For the second part of the trip, we have S0B = D is the position where the
train begins to slow, SB = L is the final position where the train arrives. The
initial speed v0B = vA in which the train is when it starts to
brake, equals the final speed of the first part of the motion. The train deceleration is
αB = b and tB the time interval that the train slows down
to stop, substituting these values in the expression (I)
\[
\begin{gather}
S_{B}=S_{0 B}+v_{0 B}t_{B}+\frac{\alpha_{B}}{2}t_{B}^{2}\\[5pt]
L=D+at_{A}t_{B}-\frac{b}{2}t_{B}^{2} \tag{V}
\end{gather}
\]
For speed, we have that the train stops when arrives at station B, the final speed will be
vB = 0, the initial speed v0B = vA will be
given by the expression (IV) above, and tB is the interval of time that the train slowed
until stops.
\[
\begin{gather}
v_{B}=v_{0 B}+\alpha_{B}t_{B}\\[5pt]
0=at_{A}-bt_{B}\\[5pt]
at_{A}=bt_{B} \tag{VI}
\end{gather}
\]
Solving expression (VI) for tB, we have the interval of time when the train slow
\[
\begin{gather}
t_{B}=\frac{a}{b}t_{A} \tag{VII}
\end{gather}
\]
substituting expressions (III) and (VII) into the expression (V)
\[
\begin{gather}
L=\frac{a}{2}t_{A}^{2}+at_{A}\frac{a}{b}t_{A}-\frac{b}{2}\left(\frac{a}{b}t_{A}\right)^{2}\\[5pt]
L=\frac{a}{2}t_{A}^{2}+\frac{a^{2}}{b}t_{A}^{2}-\frac{b}{2}\frac{a^{2}}{b^{2}}t_{A}^{2}\\[5pt]
L=\frac{a}{2}t_{A}^{2}+\frac{a^{2}}{b}t_{A}^{2}-\frac{a^{2}}{2b}t_{A}^{2}\\[5pt]
L=\frac{a}{2}t_{A}^{2}+\frac{a^{2}}{2b}t_{A}^{2}\\[5pt]
L=t_{A}^{2}\left(\frac{a}{2}+\frac{a^{2}}{2b}\right)\\[5pt]
L=t_{A}^{2}\left(\frac{ab+a^{2}}{2b}\right)
\end{gather}
\]
solving the equation for tA, we have the interval of time when the train runs the first
part of the trip
\[
\begin{gather}
t_{A}^{2}=\frac{2Lb}{a(a+b)}\\[5pt]
t_{A}=\sqrt{\frac{2Lb}{a(a+b)}}=\left[\frac{2Lb}{a(a+b)}\right]^{\frac{1}{2}} \tag{VIII}
\end{gather}
\]
Likewise, we can solve the expression (VII) for tA
\[
\begin{gather}
t_{A}=\frac{b}{a}t_{B} \tag{IX}
\end{gather}
\]
substituting this value and the expression (III) into the expression (V)
\[
\begin{gather}
L=\frac{a}{2}t_{A}^{2}+at_{A}t_{B}-\frac{b}{2}t_{B}^{2}
\end{gather}
\]
using the expression (IX) for the value of tA
\[
\begin{gather}
L=\frac{a}{2}\left(\frac{b}{a}t_{B}\right)^{2}+a\frac{b}{a}t_{B}t_{B}-\frac{b}{2}t_{B}^{2}\\[5pt]
L=\frac{a}{2}\frac{b^{2}}{a^{2}}t_{B}^{2}+a\frac{b}{a}t_{B}^{2}-\frac{b}{2}t_{B}^{2}\\[5pt]
L=\frac{b^{2}}{2a}t_{B}^{2}+bt_{B}^{2}-\frac{b}{2}t_{B}^{2}\\[5pt]
L=\frac{b^{2}}{2a}t_{B}^{2}+\frac{b}{2}t_{B}^{2}\\[5pt]
L=t_{B}^{2}\left(\frac{b^{2}}{2a}+\frac{b}{2}\right)\\[5pt]
L=t_{B}^{2}\left(\frac{b^{2}+ab}{2a}\right)
\end{gather}
\]
the interval of time until the train stops at station B will be
\[
\begin{gather}
t_{B}^{2}=\frac{2La}{b(b+a)}\\[5pt]
t_{B}=\sqrt{\frac{2La}{b(b+a)}}=\left[\frac{2La}{b(b+a)}\right]^{\frac{1}{2}} \tag{X}
\end{gather}
\]
The total time of the trip t will be given by the sum of the expressions (VIII) and (X)
\[
\begin{gather}
t=t_{A}+t_{B}\\[5pt]
t=\left[\frac{2Lb}{a(a+b)}\right]^{\frac{1}{2}}+\left[\frac{2La}{b(a+b)}\right]^{\frac{1}{2}}
\end{gather}
\]
factoring
\( \left[\dfrac{2L}{a+b}\right]^{\frac{1}{2}} \)
\[
\begin{gather}
t=\left[\frac{2L}{a+b}\right]^{\frac{1}{2}}\left[\left(\frac{b}{a}\right)^{\frac{1}{2}}+\left(\frac{a}{b}\right)^{\frac{1}{2}}\right]\\[5pt]
t=\left[\frac{2L}{a+b}\right]^{\frac{1}{2}}\;\left[\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}+\frac{a^{\frac{1}{2}}}{b^{\frac{1}{2}}}\right]\\[5pt]
t=\left[\frac{2L}{a+b}\right]^{\frac{1}{2}}\left[\frac{b^{\frac{1}{2}}b^{\frac{1}{2}}+a^{\frac{1}{2}}a^{\frac{1}{2}}}{a^{\frac{1}{2}}b^{\frac{1}{2}}}\right]\\[5pt]
t=\left[\frac{2L}{a+b}\right]^{\frac{1}{2}}\left[\frac{b+a}{a^{\frac{1}{2}}b^{\frac{1}{2}}}\right]
\end{gather}
\]
squaring the second term in the brackets
\[
\begin{gather}
t=\left[\frac{2L}{\cancel{(a+b)}}\frac{(b+a)^{\cancel{2}}}{ab}\right]^{\frac{1}{2}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t=\left[\frac{2L(b+a)}{ab}\right]^{\frac{1}{2}}}
\end{gather}
\]
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .