Solved Problem on Kinematics
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Obtain the expression for displacement as a function of time with constant acceleration from the expression of instantaneous acceleration.
Solution
The instantaneous acceleration is given by
\[ \bbox[#99CCFF,10px]
{a=\frac{dv}{dt}}
\]
We integrate this expression in dt on both sides
\[
\int {{\frac{dv}{dt'}\;dt}}=\int {{a\;dt}}
\]
as the acceleration a is constant, it is carried outside of the integral sign, and
\( \dfrac{dv}{dt}\;dt=dv \).The limits of integration are v0, the initial speed, and v(t), a speed in an instant t for dv, and t0, initial time, and t, for dt
\[
\begin{gather}
\int_{{v_{0}}}^{{v(t)}}\;dv=a\int_{{t_{0}}}^{t}\;dt\\
\left.v\;\right|_{\;v_{0}}^{\;v(t)}=a\;\left.t\;\right|_{\;t_{0}}^{\;t}\\
v(t)-v_{0}=a\left(t-t_{0}\right)
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{v(t)=v_{0}+a\left(t-t_{0}\right)}
\]
what describes the speed of a particle in motion with constant acceleration.The Instantaneous speed is given by
\[ \bbox[#99CCFF,10px]
{v=\frac{dx}{dt}}
\]
\[
\frac{dx}{dt}=v_{0}+a \left(t-t_{0}\right)
\]
we integrating this expression in dt from both sides
\[
\int {{\frac{dx}{dt}\;dt}}=\int{{\left[v_{0}+a\left(t-t_{0}\right)\right]\;dt}}
\]
in the integral, on the left-hand side,
\( \dfrac{dx}{dt}\;dt=dx \),
and on the right-hand side of the equation the integral of the sum is the sum of the integral, and as of
v0, t0 and a are constants, they are carried outside of the
integral sign.The limits of integration are x0, the initial position, and x(t), a position in an instant t for dx, and t0, initial time, and t, for dt
\[
\begin{gather}
\int_{x_{0}}^{x(t)}dx=v_{0}\int_{t_{0}}^{t}\;dt+a\int_{t_{0}}^{t}t\;dt-at_{0}\int_{t_{0}}^{t}dt\\[5pt]
\left.x\;\right|_{\;x_{0}}^{\;x(t)}=v_{0}\;\left.t\;\right|_{\;t_{0}}^{\;t}+a\;\left.\frac{t^{2}}{2}\;\right|_{\;t_{0}}^{\;t}-a t_{0}\;\left.t\;\right|_{\;t_{0}}^{\;t}\\[5pt]
x(t)-x_{0}=v_{0}\left(t-t_{0}\right)+a\left(\frac{t^{2}}{2}-\frac{t_{0}^{2}}{2}\right)-at_{0}\left(t-t_{0}\right)\\[5pt]
x(t)=x_{0}+v_{0}\left(t-t_{0}\right)+a\left(\frac{t^{2}}{2}-\frac{t_{0}^{2}}{2}-t_{0}t+t_{0}^{2}\right)\\[5pt]
x(t)=x_{0}+v_{0}\left(t-t_{0}\right)+a\left(\frac{t^{2}}{2}-t_{0}t+\frac{t_{0}^{2}}{2}\right)\\[5pt]
x(t)=x_{0}+v_{0}\left(t-t_{0}\right)+\frac{a}{2}\left(t^{2}-2t_{0}t+t_{0}^{2}\right)
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{x(t)=x_{0}+v_{0}\left(t-t_{0}\right)+\frac{a}{2}\left(t-t_{0}\right)^{2}}
\]
what describes a particle in motion in a straight line with constant acceleration.
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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 .