Solved Problem on One-dimensional Motion
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FR
PT-BR
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The table below describes the velocities of a particle moving in a given reference frame.
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| v (m/s) | 9 | 6 | 3 | 0 | -3 | -6 | -9 |
from the table find:
a) The initial speed of the particle;
b) The instant when the particle changes its direction;
c) The average acceleration of the particle between the instants 1 s and 2 s;
d) The average acceleration of the particle between the instants 5 s and 6 s.
Solution
a) From the table, we have that for t=0, the initial speed is
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{v_0=9\;\mathrm{m/s}}
\end{gather}
\]
b) At time t=2 s the velocity is positive (v>0), at time t=4 s, speed is negative (v<0), the particle changes its direction when the speed is zero at t=3 s.
c) Using the equation for the average acceleration
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\bar a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i}}
\end{gather}
\]
substituting the values in the table
\[
\begin{gather}
\bar a=\frac{3\;\mathrm{\frac{m}{s}}-6\;\mathrm{\frac{m}{s}}}{2\;\mathrm s-1\;\mathrm s}\\[5pt]
\bar a=-\frac{3\;\mathrm{\frac{m}{s}}}{1\;\mathrm s}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\bar a=-3\;\mathrm{m/s^2}}
\end{gather}
\]
d) Using the equation for the average acceleration again
\[
\begin{gather}
\bar a=\frac{-9\;\mathrm{\frac{m}{s}}-\left(-6\;\mathrm{\frac{m}{s}}\right)}{6\;\mathrm s-5\;\mathrm s}\\[5pt]
\bar a=\frac{-9\;\mathrm{\frac{m}{s}}+6\;\mathrm{\frac{m}{s}}}{1\;\mathrm s}\\[5pt]
\bar a=-\frac{3\;\mathrm{\frac{m}{s}}}{1\;\mathrm s}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\bar a=-3\;\mathrm{m/s^2}}
\end{gather}
\]
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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 .