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Furkat [3]
3 years ago
6

1) UN MOVIL A SE MUEVE DESDE UN PUNTO CON VELOCIDAD CONSTANTE DE 20m/s EN EL MISMO INSTANTE A UNA DISTANCIA DE 1200m, OTRO MOVIL

B SALE Y PERSIGUE AL MOVIL A CON VELOCIDAD CONSTANTE DE 40m/s.¿ EN QUE TIEMPO Y A QUE DISTANCIA B ALCANZA a
Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

El móvil B necesita 60 segundos para alcanzar al móvil A y le alcanza una distancia de 2400 metros con respecto al punto de referencia.

Step-by-step explanation:

Supóngase que cada movil viaja en el mismo plano y que el móvil B se localiza inicialmente en la posición x = 0\,m, mientras que el móvil A se encuentra en la posición x = 1200\,m. Ambos móviles viajan a rapidez constante. Si el móvil B alcanza al móvil A después de cierto tiempo, el sistema de ecuaciones cinemáticas es el siguiente:

Móvil A

x_{A} = 1200\,m+\left(20\,\frac{m}{s} \right)\cdot t

Móvil B

x_{B} = \left(40\,\frac{m}{s} \right)\cdot t

Donde:

x_{A}, x_{B} - Posiciones finales de cada móvil, medidas en metros.

t - Tiempo, medido en segundos.

Si x_{A} = x_{B}, el tiempo requerido por el móvil B para alcanzar al móvil A es:

1200\,m+\left(20\,\frac{m}{s} \right)\cdot t = \left(40\,\frac{m}{s} \right)t

1200\,m = \left(20\,\frac{m}{s} \right)\cdot t

t = \frac{1200\,m}{20\,\frac{m}{s} }

t = 60\,s

El móvil B necesita 60 segundos para alcanzar al móvil A.

Ahora, la distancia se obtiene por sustitución directa en cualquiera de las ecuaciones cinemáticas:

x_{B} = \left(40\,\frac{m}{s} \right)\cdot (60\,s)

x_{B} = 2400\,m

El móvil B alcanza al móvil A a una distancia de 2400 metros con respecto al punto de referencia.

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Step-by-step explanation:

x   y   rank of x     rank of y   d         d²

18   2     1                1               0          0

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