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Tecnológico Nacional de México
Instituto Tecnológico Superior del Sur del Estado de Yucatán
Ingeniería Bioquímica
Aplicaciones del Cálculo en Física (Cinemática)
Velocidad y Posición de una partícula en Función del tiempo
Eric de Jesús Puc Pech
Una partícula se mueve a lo largo del eje X con una aceleración dada por la ecuación: 𝒂 𝒕 = 𝟐𝒕𝟐
-6t, donde la
aceleración (a) esta dada en
𝒎
𝒔𝟐, y el tiempo (t) en segundos. Exprese la velocidad y la posición de la partícula en
función del tiempo si parte con una velocidad inicial de 𝟓
𝒎
𝒔
desde una posición de 2m.
Inicio del Problema
Inicio
Cuando se aplica el concepto de física en el calculo integral siempre se
busca obtener la respuesta a una variable. En este caso la velocidad
es igual a
𝒅𝒙 (𝑃𝑜𝑠𝑖𝑐𝑖ó𝑛)
𝒅𝒕 𝑡𝑖𝑒𝑚𝑝𝑜
y la aceleración es igual a
𝒅𝒗(𝑉𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑)
𝒅𝒕 𝑡𝑖𝑒𝑚𝑝𝑜
Se concluye que para resolver una
variable en relación al tiempo se
usan las ecuaciones dx= vdt y dv=
adt y con ello ya es posible
plantear las integrales.
Solución (Obteniendo la primitiva C)
1.- dv=adt, entonces se concluye que para encontrar la velocidad es:
‫׬‬ 𝑑𝑣 = ‫׬‬ 2𝑡2 − 6t dt
2.- Para obtener la función en condición al tiempo, es necesario
integrar la función:
𝑣 =
2𝑡3
3
−
6𝑡2
3
+C
3.- Para encontrar la constante de integración “C’’ tomamos de
referencia la Velocidad inicial (Vo) =5
𝑚
𝑠
y el punto inicial (t)=0.
conociendo los valores sustituimos:
5 =
2(0)3
3
− 3(0)2 + C, entonces C=5
4.- Entonces la velocidad en función al tiempo es:
𝑽 =
𝟐𝒕𝟑
𝟑
− 𝟑𝒕𝟐
+ 𝟓
5.- Una vez encontrada la función V, Ya es posible encontrar la
posición con respecto al tiempo:
න 𝑑𝑥 = න
2𝑡3
3
− 3𝑡2 + 5𝑑𝑡
6.- Integramos la función:
𝑥 =
2𝑡4
12
− 𝑡3
+ 5𝑡 + 𝑐
7.- Ahora disponemos a encontrar la constante de integración ‘’C’’
Nos indican que la posición (Xo)=
2m y el punto inicial (t)= 0,
entonces se procede a sustituir
los valores :
2 =
2(0)4
12
− 03
+ 5(0) + 𝑐
8.- Entonces C=2, En conclusión
la posición en función al tiempo
es:
𝒙 =
𝒕𝟒
𝟔
− 𝒕𝟑 + 𝟓𝒕 + 𝟐
Respuestas
𝒗 =
𝟐𝒕𝟑
𝟑
− 𝟑𝒕𝟐
+ 𝟓
𝒙 =
𝒕𝟒
𝟔
− 𝒕𝟑
+ 𝟓𝒕 + 𝟐

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Puc eric a3.3 infografia

  • 1. Tecnológico Nacional de México Instituto Tecnológico Superior del Sur del Estado de Yucatán Ingeniería Bioquímica Aplicaciones del Cálculo en Física (Cinemática) Velocidad y Posición de una partícula en Función del tiempo Eric de Jesús Puc Pech Una partícula se mueve a lo largo del eje X con una aceleración dada por la ecuación: 𝒂 𝒕 = 𝟐𝒕𝟐 -6t, donde la aceleración (a) esta dada en 𝒎 𝒔𝟐, y el tiempo (t) en segundos. Exprese la velocidad y la posición de la partícula en función del tiempo si parte con una velocidad inicial de 𝟓 𝒎 𝒔 desde una posición de 2m. Inicio del Problema Inicio Cuando se aplica el concepto de física en el calculo integral siempre se busca obtener la respuesta a una variable. En este caso la velocidad es igual a 𝒅𝒙 (𝑃𝑜𝑠𝑖𝑐𝑖ó𝑛) 𝒅𝒕 𝑡𝑖𝑒𝑚𝑝𝑜 y la aceleración es igual a 𝒅𝒗(𝑉𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑) 𝒅𝒕 𝑡𝑖𝑒𝑚𝑝𝑜 Se concluye que para resolver una variable en relación al tiempo se usan las ecuaciones dx= vdt y dv= adt y con ello ya es posible plantear las integrales. Solución (Obteniendo la primitiva C) 1.- dv=adt, entonces se concluye que para encontrar la velocidad es: ‫׬‬ 𝑑𝑣 = ‫׬‬ 2𝑡2 − 6t dt 2.- Para obtener la función en condición al tiempo, es necesario integrar la función: 𝑣 = 2𝑡3 3 − 6𝑡2 3 +C 3.- Para encontrar la constante de integración “C’’ tomamos de referencia la Velocidad inicial (Vo) =5 𝑚 𝑠 y el punto inicial (t)=0. conociendo los valores sustituimos: 5 = 2(0)3 3 − 3(0)2 + C, entonces C=5 4.- Entonces la velocidad en función al tiempo es: 𝑽 = 𝟐𝒕𝟑 𝟑 − 𝟑𝒕𝟐 + 𝟓 5.- Una vez encontrada la función V, Ya es posible encontrar la posición con respecto al tiempo: න 𝑑𝑥 = න 2𝑡3 3 − 3𝑡2 + 5𝑑𝑡 6.- Integramos la función: 𝑥 = 2𝑡4 12 − 𝑡3 + 5𝑡 + 𝑐 7.- Ahora disponemos a encontrar la constante de integración ‘’C’’ Nos indican que la posición (Xo)= 2m y el punto inicial (t)= 0, entonces se procede a sustituir los valores : 2 = 2(0)4 12 − 03 + 5(0) + 𝑐 8.- Entonces C=2, En conclusión la posición en función al tiempo es: 𝒙 = 𝒕𝟒 𝟔 − 𝒕𝟑 + 𝟓𝒕 + 𝟐 Respuestas 𝒗 = 𝟐𝒕𝟑 𝟑 − 𝟑𝒕𝟐 + 𝟓 𝒙 = 𝒕𝟒 𝟔 − 𝒕𝟑 + 𝟓𝒕 + 𝟐