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.
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
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{A_1=\frac{(B+b)h}{2}}
\end{gather}
\]
\[
\begin{gather}
A_1=\frac{[6+(6-2)]\times 2}{2} \\[5pt]
A_1=\frac{[6+4]\times 2}{2} \\[5pt]
A_1=\frac{10\times 2}{2} \\[5pt]
A_1=10
\end{gather}
\]
- Between the instants 6 s and 8 s, a trapezoid of area equal to
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{A_2=\frac{(B+b)h}{2}}
\end{gather}
\]
\[
\begin{gather}
A_2=\frac{(4+2)\times(8-6)}{2} \\[5pt]
A_2=\frac{6\times2}{2} \\[5pt]
A_2=6
\end{gather}
\]
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
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{\Delta v=|\;v_f-v_i\;|} \tag{I}
\end{gather}
\]
using the variation of the velocity found above, and the initial speed, given in the problem
(vi = v0 = 1 m/s), the velocity at t=8 s will be
\[
\begin{gather}
16\;\mathrm{m/s}=|\;v_8-1\;\mathrm{m/s}\;| \\[5pt]
16\;\mathrm{m/s}=v_8-1\;\mathrm{m/s} \\[5pt]
v_8=16\;\mathrm{m/s}+1\;\mathrm{m/s}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{v_8=17\;\mathrm{m/s}}
\end{gather}
\]
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).
- Between instants 8 s and 12 s, we have a triangle of area equal to
\[
\begin{gather}
\bbox[background:#99CCFF; padding:10px; border-radius:10px;]
{A_3=\frac{Bh}{2}}
\end{gather}
\]
\[
\begin{gather}
A_3=\frac{(12-8)\times 4}{2} \\[5pt]
A_3=\frac{4\times 4}{2} \\[5pt]
A_3=8
\end{gather}
\]
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 (vi = v8 = 17 m/s),
the expression (I) gives us
\[
\begin{gather}
8\;\mathrm{m/s}=|\;v_{12}-17\;\mathrm{m/s}\;| \\[5pt]
8\;\mathrm{m/s}=v_{12}-17\;\mathrm{m/s} \\[5pt]
v_{12}=8\;\mathrm{m/s}+17\;\mathrm{m/s}
\end{gather}
\]
\[
\begin{gather}
\bbox[background:#FFCCCC; padding:10px; border-radius:10px;]
{v_{12}=25\;\mathrm{m/s}}
\end{gather}
\]
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.
DE
ES
FR
PT-BR
