Solved Problem on One-dimensional Motion
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The velocity-time graph v = f(t) describes the motion of a body:
Find:
a) The displacement between 1 s and 9 s;
b) The average speed between 1 s and 9 s;
c) The average acceleration between 1 s and 9 s.
Solution
a) In a velocity-time graph, v = f(t), the displacement is equal to the area under the curve. The time axis is divided into units of 1 second, the velocity axis is divided into units of 1 m/s. The area of a square is side times side
\[
\begin{gather}
1\;\cancel{\mathrm s}\times 1\;\frac{\mathrm m}{\cancel{\mathrm s}}=1\;\mathrm m
\end{gather}
\]
one square of the graph represents 1 m of displacement.
Between the instants 4 s and 9 s, we have, by direct counting, 15 squares that represent 15 m of
displacement (Figure 1)
\[
\begin{gather}
A_{1}=15\;\mathrm m
\end{gather}
\]
From Figure 2, we see that between 1 s and 2 s, we have half a square which represents 0.5 m of
displacement
\[
\begin{gather}
A_{2}=0,5\;\mathrm m
\end{gather}
\]
Between 3 s and 4 s areas between velocities 0 and 1 m/s, and between 2 m/s and 3 m/s, are complete to
form a square and the displacement will be 1 m
\[
\begin{gather}
A_{3}=1\;\mathrm m
\end{gather}
\]
between the speeds 1 m/s and 2 m/s, we have half a square which represents 0.5 m of displacement
\[
\begin{gather}
A_{4}=0.5\;\mathrm m
\end{gather}
\]
Between 4 s and 6 s, the areas complete to form a square, and the displacement will be 1 m
\[
\begin{gather}
A_{5}=1\;\mathrm m
\end{gather}
\]
Between 7 s and 9 s, we have two half areas are complete to form a square, and the displacement will be 1 m
\[
\begin{gather}
A_{6}=1\;\mathrm m
\end{gather}
\]
The displacement is the total area given by the sum of the areas, found above
\[
\begin{gather}
\Delta S=A_{1}+A_{2}+A_{3}+A_{4}+A_{5}+A_{6}\\[5pt]
\Delta S=15+0.5+1+0,5+1+1
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\Delta S=19\;\mathrm m}
\end{gather}
\]
b) The average speed is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\bar v=\frac{\Delta S}{\Delta t}}
\end{gather}
\]
using the displacement found in item (a)
\[
\begin{gather}
\bar v=\frac{19}{9-1}\\[5pt]
\bar v=\frac{19}{8}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\bar v\approx 2.4\;\mathrm{m/s}}
\end{gather}
\]
c) The average acceleration is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\bar a=\frac{\Delta v}{\Delta t}}
\end{gather}
\]
From the graph, we have for t1 = 1 s the speed is v1 = 1 m/s , and for
t2 = 9 s the velocity is v2 = 2 m/s
\[
\begin{gather}
a_m=\frac{2-1}{9-1}\\[5pt]
a_m=\frac{1}{8}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_m\approx 0,1\;\mathrm{m/s^2}}
\end{gather}
\]
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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 .