Solved Problem on Kinematics
ES
FR
PT-BR
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The speed of a body is given by
\[
\begin{gather}
v=t^3-4t^2+2t+5
\end{gather}
\]
a) What is the position of the body at t = 3 s, if initially it is at x = 6 m?b) What is the acceleration of the body at t = 3 s?
Solution:
a) Initial condition
\[
\begin{gather}
x(0)=6\;\mathrm m
\end{gather}
\]
The velocity is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=\frac{dx}{dt}}
\end{gather}
\]
integrating with respect to dt on both sides of the equation
\[
\begin{gather}
\int v\;dt=\int \frac{dx}{dt}dt
\end{gather}
\]
substituting v with the function given in the problem
\[
\begin{gather}
\int dx=\int v\;dt\\[5pt]
\int dx=\int\left(t^3-4t^2+2t+5\right)\;dt\\[5pt]
x+C_1=\frac{t^{3+1}}{3+1}-4\frac{t^{2+1}}{2+1}+2\frac{t^{1+1}}{1+1}+5\frac{t^{0+1}}{0+1}+C_2\\[5pt]
x=\frac{t^4}{4}-4\frac{t^3}{3}+2\frac{t^2}{2}+5t+C_2-C_1
\end{gather}
\]
where C1 and C2 are integration constants, writing
C2−C1 as a new constant C
\[
\begin{gather}
x=\frac{1}{4}t^4-\frac{4}{3}t^3+t^2+5t+C
\end{gather}
\]
Using the initial condition, for t = 0, we have x = 6 m
\[
\begin{gather}
6=\frac{1}{4}\times 0^4-\frac{4}{3}\times 0^3+0^2+5\times 0+C\\[5pt]
C=6\;\mathrm m
\end{gather}
\]
the equation for the position as a function of time is given by
\[
\begin{gather}
x(t)=\frac{1}{4}t^4-\frac{4}{3}t^3+t^2+5t+6
\end{gather}
\]
at t = 3 s, the position will be
\[
\begin{gather}
x(3)=\frac{1}{4}\times 3^4-\frac{4}{3}\times3^3+3^2+5\times 3+6\\[5pt]
x(3)=\frac{1}{4}\times 81-\frac{4}{3}\times27+9+15+6
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{x=14.25\;\mathrm m}
\end{gather}
\]
b) The acceleration is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\end{gather}
\]
substituting v with the function given in the problem
\[
\begin{gather}
a=\frac{d\left(t^3-4t^2+2t+5\right)}{dt}\\[5pt]
a=3t^{3-1}-2\times 4t^{2-1}+1\times 2t^{1-1}-0\\[5pt]
a=3t^2-8t^1+2t^0
\end{gather}
\]
the equation for acceleration as a function of time is given by
\[
\begin{gather}
a(t)=3t^2-8t+2
\end{gather}
\]
at t = 3 s, the acceleration will be
\[
\begin{gather}
a(3)=3\times 3^2-8\times 3+2
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a=5\;\mathrm{m/s}^2}
\end{gather}
\]
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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 .