The dog in the park

Problem

A dog is at $\vec{S}_A = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ at time $t = 3\;\text{s}$, and at $\vec{S}_B = \begin{pmatrix} -4 \\ 1 \end{pmatrix}$ at time $t = 9\;\text{s}$ (positions in metres). Calculate the average velocity.

Solution

Setting up. In two dimensions the average velocity is a vector, defined as the ratio between the vector displacement and the time interval:

$\vec{v}_{\text{media}} = \frac{\vec{S}_B - \vec{S}_A}{t_B - t_A}$

Let us work by Cartesian components. The displacement is the difference of the two position vectors, and the time interval is $t_B - t_A = 9 - 3 = 6\;\text{s}$:

$\vec{v} = \frac{\begin{pmatrix} -4 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}}{9 - 3} = \frac{\begin{pmatrix} -5 \\ 3 \end{pmatrix}}{6} = \begin{pmatrix} -5/6 \\ 1/2 \end{pmatrix}\;\text{m/s}$

The two components are therefore $v_x = -\tfrac{5}{6} \approx -0{,}83\;\text{m/s}$ and $v_y = \tfrac{1}{2} = 0{,}5\;\text{m/s}$. The magnitude of the average velocity is:

$|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{\left(\tfrac{5}{6}\right)^2 + \left(\tfrac{1}{2}\right)^2} = \sqrt{\tfrac{25}{36} + \tfrac{9}{36}} = \sqrt{\tfrac{34}{36}} \approx 0{,}97\;\text{m/s}$

$\ev{\vec{v}_{\text{media}} = \begin{pmatrix} -5/6 \\ 1/2 \end{pmatrix}\;\mathrm{m/s}}$

Collegamenti

Argomenti: Kinematics · Vectors Concetti: Average velocity · Components of a vector Metodi: Decomposition into Cartesian components