Solved Problem in Kinematics

Solved Problem on Kinematics

A particle is on a plane xy, initially at rest in the position x₀ on the positive x-axis. It begins to move with constant velocities v_x, in the direction of origin, and v_y in the direction of the positive y-axis. Determine after how long the particle will be at the minimum distance of the origin and, what is this minimum distance.

Problem data:

Initial position of the particle: x₀;
Speed of the particle in direction x: v_x;
Speed of the particle in direction y: v_y.

Problem diagram:

As the components of the speed are constants the trajectory will be a straight line (Figure 1-A).

Figure 1

Along the trajectory, the particle passes successively by points P₀, P₁, P₂, P_P, P₃, and so on. The point of minimum distance to the origin will be the point PP, where the line of this point to the origin is perpendicular to the trajectory (Figure 1-B).

Solution

We have g the distance from the origin to the point of less distance, segment \( \overline{OP_{P}} \), and h the distance from the starting point to the point of less distance, segment \( \overline{x_{0}P_{P}} \). Applying the Pythagorean Theorem to the triangle ΔOP_Px₀ (in red in Figure 2)

\[ \begin{gather} x_{0}^{2}=g^{2}+h^{2}\\ h^{2}=x_{0}^{2}-g^{2} \tag{I} \end{gather} \]

Figure 2

The point P_P has coordinates (x₁, y₁), the distance traveled by the particle along the x-axis, from x₀ to x₁, will be

\[ \begin{gather} x_{0}-x_{1}=v_{x}t \tag{II} \end{gather} \]

The distance traveled by the particle along the y-axis, from O to y₁, will be

\[ y_{1}-0=v_{y}t \]

Applying the Pythagorean Theorem to the triangle Δx₁P_Px₀ (in blue in Figure 3)

\[ \begin{gather} h^{2}=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2} \tag{III} \end{gather} \]

Figure 3

The distance from the origin to point x₁, segment \( \overline{Ox_{1}} \), will be obtained from the expression (II)

\[ x_{1}=x_{0}-v_{x}t \]

Applying the Pythagorean Theorem to the triangle ΔOP_Px₁ (in blue in Figure 4)

\[ \begin{gather} g^{2}=\left(x_{0}-v_{x}t\right)^{2}+\left(v_{y}t\right)^{2} \tag{IV} \end{gather} \]

Figure 4

Equations (I), (III), and (IV) can be written as a system of three equations to three variables (g, h, and t)

\[ \left\{ \begin{array}{l} \;h^{2}=x_{0}^{2}-g^{2}\\ \;h^{2}=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2}\\ \;g^{2}=\left(x_{0}-v_{x}t\right)^{2}+\left(v_{y}t\right)^{2} \end{array} \right. \]

substituting the first equation on the second in the system

\[ \begin{gather} x_{0}^{2}-g^{2}=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2} \tag{V} \end{gather} \]

substituting the third equation of the system in the expression (V), we obtain the time interval needed to reach the minimum distance

\[ \begin{gather} x_{0}^{2}-\left[\left(x_{0}-v_{x}t\right)^{2}+\left(v_{y}t\right)^{2}\right]=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2}\\[5pt] x_{0}^{2}-\left(x_{0}-v_{x}t\right)^{2}-\left(v_{y}t\right)^{2}=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2}\\[5pt] x_{0}^{2}-\left[x_{0}^{2}-2x_{0}v_{x}t+\left(v_{x}t\right)^{2}\right]=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2}+\left(v_{y}t\right)^{2}\\[5pt] x_{0}^{2}-x_{0}^{2}+2x_{0}v_{x}t-\left(v_{x}t\right)^{2}=\left(v_{x}t\right)^{2}+2\left(v_{y}t\right)^{2}\\[5pt] 2x_{0}v_{x}t=\left(v_{x}t\right)^{2}+\left(v_{x}t\right)^{2}+2\left(v_{y}t\right)^{2}\\[5pt] 2x_{0}v_{x}t=2\left(v_{x}t\right)^{2}+2\left(v_{y}t\right)^{2}\\[5pt] x_{0}v_{x}t=\left(v_{x}t\right)^{2}+\left(v_{y}t\right)^{2}\\[5pt] v_{x}^{2}t^{2}+v_{y}^{2}t^{2}=x_{0}v_{x}t\\[5pt] t^{2}\left(v_{x}^{2}+v_{y}^{2}\right)=x_{0}v_{x}t\\[5pt] t\left(v_{x}^{2}+v_{y}^{2}\right)=x_{0}v_{x} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {t=\frac{x_{0}v_{x}}{v_{x}^{2}+v_{y}^{2}}} \]

From expression (V), we get a minimum distance

\[ \begin{gather} g^{2}=x_{0}^{2}-v_{x}^{2}t^{2}-v_{y}^{2}t^{2}\\ g^{2}=x_{0}^{2}-t^{2}\left(v_{x}^{2}+v_{y}^{2}\right) \end{gather} \]

substituting the time value found above

\[ \begin{gather} g^{2}=x_{0}^{2}-\left(\frac{x_{0}v_{x}}{v_{x}^{2}+v_{y}^{2}}\right)^{2}\left(v_{x}^{2}+v_{y}^{2}\right)\\[5pt] g^{2}=x_{0}^{2}-\frac{x_{0}^{2}v_{x}^{2}}{\left(v_{x}^{2}+v_{y}^{2}\right)^{2}}\left(v_{x}^{2}+v_{y}^{2}\right)\\[5pt] g^{2}=x_{0}^{2}-\frac{x_{0}^{2}v_{x}^{2}}{v_{x}^{2}+v_{y}^{2}}\\[5pt] g^{2}=\frac{x_{0}^{2}\left(v_{x}^{2}+v_{y}^{2}\right)-x_{0}^{2}v_{x}^{2}}{v_{x}^{2}+v_{y}^{2}}\\[5pt] g^{2}=\frac{x_{0}^{2}v_{x}^{2}+x_{0}^{2}v_{y}^{2}-x_{0}^{2}v_{x}^{2}}{v_{x}^{2}+v_{y}^{2}}\\[5pt] g^{2}=\frac{x_{0}^{2}v_{x}^{2}+x_{0}^{2}v_{y}^{2}-x_{0}^{2}v_{x}^{2}}{v_{x}^{2}+v_{y}^{2}}\\[5pt] g^{2}=\frac{x_{0}^{2}v_{y}^{2}}{v_{x}^{2}+v_{y}^{2}}\\[5pt] g=\sqrt{\frac{x_{0}^{2}v_{y}^{2}}{v_{x}^{2}+v_{y}^{2}}\;} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {g=\frac{x_{0}v_{y}}{\sqrt{v_{x}^{2}+v_{y}^{2}}\;}} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .