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

Figure 1

We can divide the graph into the following areas (figure 1)
  • Between the instants, 1 s and 2 s, a triangle of area equal to
\[ \begin{gather} \bbox[#99CCFF,10px] {A_1=\frac{Bh}{2}} \end{gather} \]
\[ \begin{gather} A_1=\frac{(2-1)\times 1}{2}\\[5pt] A_1=\frac{1\times 1}{2}\\[5pt] A_1=\frac{1}{2}\\[5pt] A_1=0.5 \end{gather} \]
  • Between the instants, 2 s and 3 s, the speed is zero, the body remained at rest, there was no displacement.
  • Between the instants, 3 s and 4 s, a triangle of area equal to
\[ \begin{gather} \bbox[#99CCFF,10px] {A_2=\frac{Bh}{2}} \end{gather} \]
\[ \begin{gather} A_2=\frac{(4-3)\times 3}{2}\\[5pt] A_2=\frac{1\times 3}{2}\\[5pt] A_2=\frac{3}{2}\\[5pt] A_2=1.5 \end{gather} \]
  • Between the instants, 4 s and 6 s, a trapezoid of area equal to
\[ \begin{gather} \bbox[#99CCFF,10px] {A_3=\frac{(B+b)h}{2}} \end{gather} \]
\[ \begin{gather} A_3=\frac{(4+3)\times(6-4)}{2}\\[5pt] A_3=\frac{7\times 2}{2}\\[5pt] A_3=7 \end{gather} \]
  • Between the instants, 6 s and 7 s, a rectangle of area equal to
\[ \begin{gather} \bbox[#99CCFF,10px] {A_4=L_1 L_2} \end{gather} \]
\[ \begin{gather} A_4=4\times (7-6)\\[5pt] A_4=4\times 1\\[5pt] A_4=4 \end{gather} \]
  • Between the instants, 7 s and 9 s, a trapezoid of area equal to
\[ \begin{gather} \bbox[#99CCFF,10px] {A_5=\frac{(B+b)h}{2}} \end{gather} \]
\[ \begin{gather} A_5=\frac{(4+2)\times(9-7)}{2}\\[5pt] A_5=\frac{6\times 2}{2}\\[5pt] A_5=6 \end{gather} \]
The total area will be the sum of the areas found above
\[ \begin{gather} \Delta S=A_{1}+A_{2}+A_{3}+A_{4}+A_{5}\\[5pt] \Delta S=0.5+1.5+7+4+6 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta S=19\;\mathrm m} \end{gather} \]

b) The average speed is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {v_m=\frac{\Delta S}{\Delta t}} \end{gather} \]
using the displacement 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
\[ \begin{gather} \bar a=\frac{2-1}{9-1}\\[5pt] \bar a=\frac{1}{8} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\bar a\approx 0,1\;\mathrm{m/s^2}} \end{gather} \]
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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 .