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Marizza181
29 days ago
14

Un cuerpo se mueve en línea recta segun la ecuación x=10+20t-4.9t2 (x está expresado en metros y t en segundos). ¿Cuál es la lon

gitud del camino recorrido al cabo de 3.0 segundos? Porfa es para mañana
Physics
2 answers:
serg [3.5K]29 days ago
6 0

Respuesta:

La fórmula que estamos considerando es:

X = 10m + 20m/s*t - 4.9m/s^2*t^2

donde x representa la ubicación de un objeto en metros y t en segundos.

Ahora, debemos calcular la distancia que se desplaza el objeto en 3 segundos.

Esto se puede determinar así:

X(3) - X(0) = 10m + 20m/s*3 - 4.9m/s^2*3^2 - 10m = 15.9m

Por tanto, el objeto se mueve 15.9 metros en los primeros 3 segundos.

ahora en español.

La ecuación que tenemos es:

X = 10m + 20m / s * t - 4.9m / s ^ 2 * t ^ 2

donde x es la posición de un objeto en metros yt está en segundos.

Ahora, queremos saber la distancia que se mueve el objeto en 3 segundos.

Esto se puede calcular como:

X (3) - X (0) = 10m + 20m / s * 3 - 4.9m / s ^ 2 * 3 ^ 2 - 10m = 15.9m

Entonces el objeto se mueve 15.9 metros en los primeros 3 segundos.

Keith_Richards [3.2K]29 days ago
4 0

Respuesta:

La distancia recorrida es de 25.9 [m]

Explicación:

Se sustituye el valor del tiempo en segundos en la fórmula de desplazamiento

x=10+20*(3) - 4.9*(3)^2

x= 25.9 [metros]

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