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
\( \dfrac{dv}{dt}\;dt=dv \)
\[
\begin{gather}
\int {{dv}}=a\int{{dt}}\\
v(t)+C_{1}=at+C_{2}\\
v(t)=at+C_{2}-C_{1}
\end{gather}
\]
C1 and C2 are integration constants that can be defined as a function of
a new constant C = C2 − C1
\[
\begin{gather}
v(t)=at+C \tag{I}
\end{gather}
\]
assuming that in the initial instant, t0, the particle is with the initial speed,
v0, we have the initial condition v(t0) = v0,
substituting in the expression (I)
\[
\begin{gather}
v(t_{0})=at_{0}+C\\
v_{0}=at_{0}+C\\
C=v_{0}-at_{0} \tag{II}
\end{gather}
\]
substituting the expression (II) into expression (I)
\[
v(t)=at+v_{0}-at_{0}
\]
\[ \bbox[#FFCCCC,10px]
{v(t)=v_{0}+a\left(t-t_{0}\right)}
\]
what describes the speed of a particle in motion with constant acceleration.The Instantaneous speed is given by
\[ \bbox[#99CCFF,10px]
{v=\frac{dx}{dt}}
\]
substituting this expression in the result obtained above
\[
\frac{dx}{dt}=v_{0}+a\left(t-t_{0}\right)
\]
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
\[
\begin{gather}
\int {{dx}}=v_{0}\int {{dt}}+a\int {{t\;dt}}-at_{0}\int{{dt}}\\
x(t)+C_{1}=v_{0}t+C_{2}+a\frac{t^{2}}{2}+C_{3}-at_{0}t+C_{4}\\
x(t)=v_{0}t+a\frac{t^{2}}{2}-at_{0}t+C_{2}+C_{3}+C_{4}-C_{1}
\end{gather}
\]
C1, C2, C3 and C4 are integration
constants that can be defined as function of a new constant
C = C2+C3+C4−C1
\[
x(t)=v_{0}t+\frac{a}{2}\left(t^{2}-2t_{0}t\right)+C
\]
on the right-hand side, in the parentheses, we add and subtract t02
\[
\begin{gather}
x(t)=v_{0}t+\frac{a}{2}\left(t^{2}-2t_{0}t+t_{0}^{2}-t_{0}^{2}\right)+C\\
x(t)=v_{0}t+\frac{a}{2}\left(t-t_{0}\right)^{2}-\frac{a}{2}t_{0}^{2}+C \tag{III}
\end{gather}
\]
assuming that in the initial instant t0, the particle is in the initial position,
x0, we have the initial condition x(t0) = x0,
substituting in the expression (III)
\[
\begin{gather}
x(t_{0})=v_{0}t_{0}+\frac{a}{2}\left(t_{0}-t_{0}\right)^{2}-\frac{a}{2}t_{0}^{2}+C\\
C=x(t_{0})-v_{0}t_{0}-\frac{a}{2}.0^{2}+\frac{a}{2}t_{0}^{2}\\
C=x_{0}-v_{0}t_{0}+\frac{a}{2}t_{0}^{2} \tag{IV}
\end{gather}
\]
substituting the expression (IV) into expression (III)
\[
x(t)=v_{0}t+\frac{a}{2}\left(t-t_{0}\right)^{2}-\frac{a}{2}t_{0}^{2}+x_{0}-v_{0}t_{0}+\frac{a}{2}t_{0}^{2}
\]
\[ \bbox[#FFCCCC,10px]
{x(t)=x_{0}+v_{0}\left(t-t_{0}\right)+\frac{a}{2}\left(t-t_{0}\right)^{2}}
\]
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 .