Solved Problem on One-dimensional Motion

A particle starts a straight motion with an initial speed of 1 m/s. Given the acceleration-time graph from the start of the movement. Find:

a) The velocity at t=8 s;
b) The velocity at t=12 s;
c) The velocity at t=14 s;
d) Between, which time interval the speed decreases.

Problem Data:

Initial velocity of particle: v₀ = 1 m/s.

Solution

a) In an acceleration-time graph, a = f(t), the change in speed is equal to the area under the curve.

Figure 1

We can divide the graph into the following areas (figure 1):

Between the instants 0 s and 6 s, a trapezoid of area equal to

\[ \begin{gather} \bbox[#99CCFF,10px] {A_1=\frac{(B+b)h}{2}} \end{gather} \]

\[ \begin{gather} A_1=\frac{[6+(6-2)]\times 2}{2}\\[5pt] A_1=\frac{[6+4]\times 2}{2}\\[5pt] A_1=\frac{10\times 2}{2}\\[5pt] A_1=10 \end{gather} \]

Between the instants 6 s and 8 s, a trapezoid of area equal to

\[ \begin{gather} \bbox[#99CCFF,10px] {A_2=\frac{(B+b)h}{2}} \end{gather} \]

\[ \begin{gather} A_2=\frac{(4+2)\times(8-6)}{2}\\[5pt] A_2=\frac{6\times2}{2}\\[5pt] A_2=6 \end{gather} \]

The variation in velocity will be the total area given by the sum of the areas found.

\[ \begin{gather} \Delta v=A_1+A_2\\[5pt] \Delta v=10+6\\[5pt] \Delta v=16\;\mathrm{m/s} \end{gather} \]

From the definition of change of speed

\[ \begin{gather} \bbox[#99CCFF,10px] {\Delta v=|\;v_f-v_i\;|} \tag{I} \end{gather} \]

using the variation of the velocity found above, and the initial speed, given in the problem (v_i = v₀ = 1 m/s), the velocity at t=8 s will be

\[ \begin{gather} 16\;\mathrm{m/s}=|\;v_8-1\;\mathrm{m/s}\;|\\[5pt] 16\;\mathrm{m/s}=v_8-1\;\mathrm{m/s}\\[5pt] v_8=16\;\mathrm{m/s}+1\;\mathrm{m/s} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_8=17\;\mathrm{m/s}} \end{gather} \]

b) To find the velocity at t=12 s, we must find the change in velocity between t=8 s and t=12 s (Figure 2).

Figure 2

Between instants 8 s and 12 s, we have a triangle of area equal to

\[ \begin{gather} \bbox[#99CCFF,10px] {A_3=\frac{Bh}{2}} \end{gather} \]

\[ \begin{gather} A_3=\frac{(12-8)\times 4}{2}\\[5pt] A_3=\frac{4\times 4}{2}\\[5pt] A_3=8 \end{gather} \]

Since this area is equal to the velocity variation (Δv), and the velocity found in item (a), to be the initial velocity of this part of the motion (v_i = v₈ = 17 m/s), the expression (I) gives us

\[ \begin{gather} 8\;\mathrm{m/s}=|\;v_{12}-17\;\mathrm{m/s}\;|\\[5pt] 8\;\mathrm{m/s}=v_{12}-17\;\mathrm{m/s}\\[5pt] v_{12}=8\;\mathrm{m/s}+17\;\mathrm{m/s} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_{12}=25\;\mathrm{m/s}} \end{gather} \]

c) Between 12 s and 14 s , the acceleration is zero (a = 0), there is no change in the particle velocity, its velocity remains constant.

d) The speed does not decrease at any time interval.

Note: between instants 8 s and 12 s, the graph shows a line where acceleration decreases over time. However this does not mean that the speed decreases, the line is above the abscissa axis (time axis), the acceleration has a positive value (a>0), and the velocity increases at a lower rate. For the velocity to decrease, the line would need to be below the abscissa axis, the acceleration would have a negative value (a<0), and the object would be decelerating.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .