【Rpta.】La rapidez de la persona es de 21 km/h.
[tex]{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}[/tex]
La ecuación escalar que utilizaremos para determinar la rapidez en un movimiento rectilíneo uniforme(MRU) es:
[tex]\boxed{\boldsymbol{\mathsf{v=\dfrac{d}{t}}}} \hspace{25pt} \mathsf{Donde}\hspace{20pt} \overset{\displaystyle \nearrow \overset{\displaystyle \mathsf{v:rapidez}}{\vphantom{A}}}{\vphantom{\frac{a}{a}}} \kern-55pt\rightarrow\mathsf{d:distancia}\kern-66pt\underset{\displaystyle\searrow \underset{\displaystyle \mathsf{t:tiempo}}{}}{}[/tex]
Extraemos los datos del problema:
[tex]\mathsf{\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:d = 42\: km}[/tex]
[tex]\mathsf{\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\: t = 2\: h}[/tex]
Reemplazamos estos valores en la ecuación escalar anterior.
[tex]\mathsf{\:\:\:\:\:\:\:\:v = \dfrac{\mathsf{d}}{\mathsf{t}}}\\\\\\\mathsf{\:\:\:\:\:v = \dfrac{\mathsf{42\:km}}{\mathsf{2\:h}}}\\\\\\\mathsf{\:\:\:v = \dfrac{\mathsf{42\!\!\!\!\!\!\!\dfrac{\hspace{0.4cm}}{~}^{21}\:km}}{\mathsf{2\!\!\!\!\!\!\!\dfrac{\hspace{0.4cm}}{~}_{1}\:h}}}\\\\\\\mathsf{\boxed{\boxed{\boldsymbol{\mathsf{v = 21\:km/h}}}}}[/tex]
✠ Tareas similares
➫ https://brainly.lat/tarea/19397837
➫ https://brainly.lat/tarea/19012248
➫ https://brainly.lat/tarea/16460446
[tex]\mathsf{\mathsf{\above 3pt \phantom{aa}\overset{\displaystyle \fbox{I\kern-3pt R}}{}\hspace{4 pt}\displaystyle \fbox{C\kern-6.5pt O}\hspace{4 pt}\overset{\displaystyle\fbox{C\kern-6.5pt G}}{} \hspace{4 pt} \displaystyle \fbox{I\kern-3pt H} \hspace{4pt}\overset{\displaystyle\fbox{I\kern-3pt E}}{} \hspace{4pt}\displaystyle \fbox{I\kern-3pt R} \phantom{aa}} \above 3pt}[/tex]
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【Rpta.】La rapidez de la persona es de 21 km/h.
[tex]{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}[/tex]
La ecuación escalar que utilizaremos para determinar la rapidez en un movimiento rectilíneo uniforme(MRU) es:
[tex]\boxed{\boldsymbol{\mathsf{v=\dfrac{d}{t}}}} \hspace{25pt} \mathsf{Donde}\hspace{20pt} \overset{\displaystyle \nearrow \overset{\displaystyle \mathsf{v:rapidez}}{\vphantom{A}}}{\vphantom{\frac{a}{a}}} \kern-55pt\rightarrow\mathsf{d:distancia}\kern-66pt\underset{\displaystyle\searrow \underset{\displaystyle \mathsf{t:tiempo}}{}}{}[/tex]
Extraemos los datos del problema:
[tex]\mathsf{\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\:d = 42\: km}[/tex]
[tex]\mathsf{\boldsymbol{\bigcirc \kern-8pt \triangleright}\:\:\:\: t = 2\: h}[/tex]
Reemplazamos estos valores en la ecuación escalar anterior.
[tex]\mathsf{\:\:\:\:\:\:\:\:v = \dfrac{\mathsf{d}}{\mathsf{t}}}\\\\\\\mathsf{\:\:\:\:\:v = \dfrac{\mathsf{42\:km}}{\mathsf{2\:h}}}\\\\\\\mathsf{\:\:\:v = \dfrac{\mathsf{42\!\!\!\!\!\!\!\dfrac{\hspace{0.4cm}}{~}^{21}\:km}}{\mathsf{2\!\!\!\!\!\!\!\dfrac{\hspace{0.4cm}}{~}_{1}\:h}}}\\\\\\\mathsf{\boxed{\boxed{\boldsymbol{\mathsf{v = 21\:km/h}}}}}[/tex]
✠ Tareas similares
➫ https://brainly.lat/tarea/19397837
➫ https://brainly.lat/tarea/19012248
➫ https://brainly.lat/tarea/16460446
[tex]\mathsf{\mathsf{\above 3pt \phantom{aa}\overset{\displaystyle \fbox{I\kern-3pt R}}{}\hspace{4 pt}\displaystyle \fbox{C\kern-6.5pt O}\hspace{4 pt}\overset{\displaystyle\fbox{C\kern-6.5pt G}}{} \hspace{4 pt} \displaystyle \fbox{I\kern-3pt H} \hspace{4pt}\overset{\displaystyle\fbox{I\kern-3pt E}}{} \hspace{4pt}\displaystyle \fbox{I\kern-3pt R} \phantom{aa}} \above 3pt}[/tex]