Solved Problem on Kinematics
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Two bodies are in a straight line with constant acceleration on the same trajectory, their motions are described by the equations
\[
\begin{gather}
\left.
\begin{array}{l}
x_{1}=2t-\dfrac{1}{2}t^{2}\\[5pt]
x_{2}=10-3t+\dfrac{3}{2}t^{2}
\end{array}
\right.
\qquad\text{(}\mathit{S.I.}\text{ units)}
\end{gather}
\]
Determine:a) The position where they meet;
b) The instant of time that the distance between the two bodies is the smallest and the distance between them;
c) The instant of time in which the speeds of the bodies change direction and their positions.
Solution
a) When bodies meet their positions should be equal
\[
\begin{gather}
x_{1}=x_{2}\\
2t-\frac{1}{2}t^{2}=10-3t+\frac{3}{2}t^{2}\\
10-3t+\frac{3}{2}t^{2}-2t+\frac{1}{2}t^{2}=0\\
10-5t+2t^{2}=0\\[10pt]
\Delta=(-5)^{2}-4\times 2\times 10\\
\Delta =25-80\\
\Delta =-55
\end{gather}
\]
As Δ < 0 this means that no x satisfies the two equations at the same time,
the bodies do not meet.
b) At any point, the distance between bodies is given by the difference in their positions
\[
\begin{gather}
x=x_{2}-x_{1}\\
x=10-3t+\frac{3}{2}t^{2}-\left(2t-\frac{1}{2}t^{2}\right)\\
x=10-3t+\frac{3}{2}t^{2}-2t+\frac{1}{2}t^{2}\\
x=10-5t+2t^{2} \tag{I}
\end{gather}
\]
To determine the minimum distance, we take the derivative of this function with respect to time, and
equal to zero
\[
\begin{gather}
\frac{dx}{dt}=-5+4t=0 \tag{II}\\
4t=5
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{t=\frac{5}{4}=1.25\;\text{s}}
\]
To verify if this is the minimum point of the function, we take the second derivative, deriving the
expression (II)
\[
\frac{d^{2}x}{dt^{2}}=4 > 0
\]
as the second derivative is greater than zero the instant found represents a minimum point of the
function.Substituting the instant of time calculated in the expression (I), we find the minimum xmin distance between the bodies
\[
\begin{gather}
x_{min}=10-5\times \frac{5}{4}+2\times \left(\frac{5}{4}\right)^{2}\\
x_{min}=10-\frac{25}{4}+2\times \frac{25}{16}\\
x_{min}=10-\frac{25}{4}+\frac{50}{16}\\
x_{min}=\frac{160-100+50}{16}\\
x_{min}=\frac{110}{16}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{x_{min}=6.875\;\text{m}}
\]
c) To find the expressions of velocities as a function of time for bodies of 1 and 2, we take the derivative of their expressions with respect to time, for the body 1
\[
\frac{dx_{1}}{dt}=v_{1}=2-t
\]
when the speed changes direction it equals zero, v1 = 0
\[
0=2-t
\]
\[ \bbox[#FFCCCC,10px]
{t=2\;\text{s}}
\]
substituting this value in the expression for x1, the position will be
\[
\begin{gather}
x_{1}=2\times 2-\frac{1}{2}\times 2^{2}\\
x_{1}=4-\frac{4}{2}\\
x_{1}=4-2
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{x_{1}=2\;\text{m}}
\]
For body 2
\[
\frac{dx_{2}}{dt}=v_{2}=-3+3t
\]
when the speed changes direction it equals zero, v2 = 0
\[
\begin{gather}
0=-3+3t\\
3t=3\\
t=\frac{3}{3}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{t=1\;\text{s}}
\]
substituting this value in the expression for x2, the position will be
\[
\begin{gather}
x_{2}=10-3.1+\frac{3}{2}\times 1^{2}\\
x_{2}=10-3+\frac{3}{2}\\
x_{2}=\frac{20-6+3}{2}\\
x_{2}=\frac{17}{2}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{x_{2}=8.5\;\text{m}}
\]
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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 .