【Rpta.】La distancia que recorre el automóvil es de 900 metros.
[tex]{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}[/tex]
La ecuación escalar que utilizaremos para determinar la distancia en un movimiento rectilíneo uniformemente variado(MRUV) es:
[tex]\boxed{\boldsymbol{\mathsf{d=\left(\dfrac{v_{o}+v_{f}}{2}\right)t}}} \hspace{20pt} \mathsf{Donde} \hspace{10pt}\overset{\displaystyle\overset{\displaystyle \mathsf{\rightarrow v_f:rapidez\:final\kern12pt \rightarrow d:distancia}}{\vphantom{A}}}{\vphantom{\frac{a}{a}}}\kern-156pt\underset{\displaystyle \underset{\displaystyle \mathsf{\rightarrow v_o:rapidez\:inicial\kern5pt\rightarrow t:tiempo}}{}}{}[/tex]
Extraemos los datos del enunciado
[tex]\mathsf{\blacktriangleright v_o=80\:m/s}[/tex] [tex]\mathsf{\blacktriangleright v_f=100\:m/s}[/tex] [tex]\mathsf{\blacktriangleright t=10\:s}[/tex]
Reemplazamos estos valores en la ecuación escalar
[tex]\mathsf{\:\:\:\:d = \left(\dfrac{v_{o} + v_{f}}{2}\right)t}\\\\\\\mathsf{d = \left(\dfrac{80 + 100}{2}\right)(10)}\\\\\\\mathsf{\:\:\:\:d = \left(\dfrac{180}{2}\right)(10)}\\\\\\\mathsf{\:\:\:\:\:\:d = (90)(10)}\\\\\\\mathsf{\:\:\:\boxed{\boxed{\boldsymbol{\mathsf{d = 900\:m}}}}}[/tex]
✠ Tareas similares
➫ https://brainly.lat/tarea/19319718
[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 distancia que recorre el automóvil es de 900 metros.
[tex]{\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedimiento}}{\hspace{50pt}\above 1.2pt}[/tex]
La ecuación escalar que utilizaremos para determinar la distancia en un movimiento rectilíneo uniformemente variado(MRUV) es:
[tex]\boxed{\boldsymbol{\mathsf{d=\left(\dfrac{v_{o}+v_{f}}{2}\right)t}}} \hspace{20pt} \mathsf{Donde} \hspace{10pt}\overset{\displaystyle\overset{\displaystyle \mathsf{\rightarrow v_f:rapidez\:final\kern12pt \rightarrow d:distancia}}{\vphantom{A}}}{\vphantom{\frac{a}{a}}}\kern-156pt\underset{\displaystyle \underset{\displaystyle \mathsf{\rightarrow v_o:rapidez\:inicial\kern5pt\rightarrow t:tiempo}}{}}{}[/tex]
Extraemos los datos del enunciado
[tex]\mathsf{\blacktriangleright v_o=80\:m/s}[/tex] [tex]\mathsf{\blacktriangleright v_f=100\:m/s}[/tex] [tex]\mathsf{\blacktriangleright t=10\:s}[/tex]
Reemplazamos estos valores en la ecuación escalar
[tex]\mathsf{\:\:\:\:d = \left(\dfrac{v_{o} + v_{f}}{2}\right)t}\\\\\\\mathsf{d = \left(\dfrac{80 + 100}{2}\right)(10)}\\\\\\\mathsf{\:\:\:\:d = \left(\dfrac{180}{2}\right)(10)}\\\\\\\mathsf{\:\:\:\:\:\:d = (90)(10)}\\\\\\\mathsf{\:\:\:\boxed{\boxed{\boldsymbol{\mathsf{d = 900\:m}}}}}[/tex]
✠ Tareas similares
➫ https://brainly.lat/tarea/19319718
[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]