【Rpta.】La distancia que recorre el dragster a los 1, 2 y 3 segundos es 3.7, 14.8 y 33.3 metros respectivamente.
[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 = v_{o}t \pm \dfrac{at^2}{2}}}} \hspace{20pt} \mathsf{Donde} \hspace{10pt}\overset{\displaystyle\overset{\displaystyle \mathsf{\rightarrow t:tiempo\kern40pt \rightarrow a:aceleraci\acute{o}n}}{\vphantom{A}}}{\vphantom{\frac{a}{a}}}\kern-168pt\underset{\displaystyle \underset{\displaystyle \mathsf{\rightarrow v_o:rapidez\:inicial\kern8pt\rightarrow d:distancia}}{}}{}[/tex]
El signo positivo se utiliza cuando el móvil acelera, mientra que el negativo cuando desacelera.
Datos
[tex]\mathsf{\blacktriangleright v_o=0\:m/s}[/tex] [tex]\mathsf{\blacktriangleright a=7.4\:m/s^2}[/tex] [tex]\mathsf{\blacktriangleright t=1\:s}[/tex]
Reemplazamos estos valores en la ecuación escalar
[tex]\mathsf{\:\:\:\:\:\:\:d = v_{o}t + \dfrac{at^2}{2}}\\\\\\\mathsf{d = (0)(1) + \dfrac{(7.4)(1)^2}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:d = \dfrac{(7.4)(1)}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:\:\:\:d = \dfrac{7.4}{2}}\\\\\\\mathsf{\:\:\:\:\:\boxed{\boldsymbol{\boxed{\mathsf{d = 3.7\:m}}}}}[/tex]
[tex]\mathsf{\blacktriangleright v_o=0\:m/s}[/tex] [tex]\mathsf{\blacktriangleright a=7.4\:m/s^2}[/tex] [tex]\mathsf{\blacktriangleright t=2\:s}[/tex]
[tex]\mathsf{\:\:\:\:\:\:\:d = v_{o}t + \dfrac{at^2}{2}}\\\\\\\mathsf{d = (0)(1) + \dfrac{(7.4)(2)^2}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:d = \dfrac{(7.4)(4)}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:\:\:\:d = \dfrac{29.6}{2}}\\\\\\\mathsf{\:\:\:\:\:\boxed{\boldsymbol{\boxed{\mathsf{d = 14.8\:m}}}}}[/tex]
[tex]\mathsf{\blacktriangleright v_o=0\:m/s}[/tex] [tex]\mathsf{\blacktriangleright a=7.4\:m/s^2}[/tex] [tex]\mathsf{\blacktriangleright t=3\:s}[/tex]
[tex]\mathsf{\:\:\:\:\:\:\:d = v_{o}t + \dfrac{at^2}{2}}\\\\\\\mathsf{d = (0)(1) + \dfrac{(7.4)(3)^2}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:d = \dfrac{(7.4)(9)}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:\:\:\:d = \dfrac{66.6}{2}}\\\\\\\mathsf{\:\:\:\:\:\boxed{\boldsymbol{\boxed{\mathsf{d = 33.3\:m}}}}}[/tex]
✠ Tareas similares
➫ https://brainly.lat/tarea/53879354
➫ https://brainly.lat/tarea/50248147
➫ https://brainly.lat/tarea/20041594
[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]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
【Rpta.】La distancia que recorre el dragster a los 1, 2 y 3 segundos es 3.7, 14.8 y 33.3 metros respectivamente.
[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 = v_{o}t \pm \dfrac{at^2}{2}}}} \hspace{20pt} \mathsf{Donde} \hspace{10pt}\overset{\displaystyle\overset{\displaystyle \mathsf{\rightarrow t:tiempo\kern40pt \rightarrow a:aceleraci\acute{o}n}}{\vphantom{A}}}{\vphantom{\frac{a}{a}}}\kern-168pt\underset{\displaystyle \underset{\displaystyle \mathsf{\rightarrow v_o:rapidez\:inicial\kern8pt\rightarrow d:distancia}}{}}{}[/tex]
El signo positivo se utiliza cuando el móvil acelera, mientra que el negativo cuando desacelera.
✅ Distancia que recorre para 1 segundo.
Datos
[tex]\mathsf{\blacktriangleright v_o=0\:m/s}[/tex] [tex]\mathsf{\blacktriangleright a=7.4\:m/s^2}[/tex] [tex]\mathsf{\blacktriangleright t=1\:s}[/tex]
Reemplazamos estos valores en la ecuación escalar
[tex]\mathsf{\:\:\:\:\:\:\:d = v_{o}t + \dfrac{at^2}{2}}\\\\\\\mathsf{d = (0)(1) + \dfrac{(7.4)(1)^2}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:d = \dfrac{(7.4)(1)}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:\:\:\:d = \dfrac{7.4}{2}}\\\\\\\mathsf{\:\:\:\:\:\boxed{\boldsymbol{\boxed{\mathsf{d = 3.7\:m}}}}}[/tex]
✅ Distancia que recorre para 2 segundos.
Datos
[tex]\mathsf{\blacktriangleright v_o=0\:m/s}[/tex] [tex]\mathsf{\blacktriangleright a=7.4\:m/s^2}[/tex] [tex]\mathsf{\blacktriangleright t=2\:s}[/tex]
Reemplazamos estos valores en la ecuación escalar
[tex]\mathsf{\:\:\:\:\:\:\:d = v_{o}t + \dfrac{at^2}{2}}\\\\\\\mathsf{d = (0)(1) + \dfrac{(7.4)(2)^2}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:d = \dfrac{(7.4)(4)}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:\:\:\:d = \dfrac{29.6}{2}}\\\\\\\mathsf{\:\:\:\:\:\boxed{\boldsymbol{\boxed{\mathsf{d = 14.8\:m}}}}}[/tex]
✅ Distancia que recorre para 3 segundos.
Datos
[tex]\mathsf{\blacktriangleright v_o=0\:m/s}[/tex] [tex]\mathsf{\blacktriangleright a=7.4\:m/s^2}[/tex] [tex]\mathsf{\blacktriangleright t=3\:s}[/tex]
Reemplazamos estos valores en la ecuación escalar
[tex]\mathsf{\:\:\:\:\:\:\:d = v_{o}t + \dfrac{at^2}{2}}\\\\\\\mathsf{d = (0)(1) + \dfrac{(7.4)(3)^2}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:d = \dfrac{(7.4)(9)}{2}}\\\\\\\mathsf{\:\:\:\:\:\:\:\:\:\:\:d = \dfrac{66.6}{2}}\\\\\\\mathsf{\:\:\:\:\:\boxed{\boldsymbol{\boxed{\mathsf{d = 33.3\:m}}}}}[/tex]
✠ Tareas similares
➫ https://brainly.lat/tarea/53879354
➫ https://brainly.lat/tarea/50248147
➫ https://brainly.lat/tarea/20041594
[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]