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Galina-37 [17]
3 years ago
7

C

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
5 0

Answer:

x3

Step-by-step explanation:

because we have to exchange y and x

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About 3,000 books because 105 * 28 = 2940
and 2940 rounded to the nearest thousand is 3000. Hope this helps.
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Escribe la posición del móvil si el diámetro de la trayectoria es de 10 m y la distancia recorrida es de 190 m
lions [1.4K]

Step-by-step explanation:

La posici´on de una part´ıcula que se mueve unidimensionalmente esta definida por la ecuaci´on:

x(t) = 2t

3 − 15t

2 + 24t + 4 donde 0x

0 y

0

t

0

se expresan en metros y segundos respectivamente. Determine:

a. ¿Cu´ando la velocidad es cero?

b. La posici´on y la distancia total recorrida cuando la aceleraci´on es cero.

Soluci´on:

a. Recordemos que:

v(t) =

dx

dt =

d

dt(2t

3 − 15t

2 + 24t + 4) = 6t

2 − 30t + 24

Sea t

0

el tiempo en que la velocidad se anula, entonces v(t

0

) = 0.

De este modo:

0 = v(t

0

) = 6(t

0

)

2 − 30(t

0

) + 24 = 6[(t

0

)

2 − 5(t

0

) + 4] = 6[(t

0

) − 4][(t

0

) − 1]

As´ı tenemos que:

t

0

1 = 4, t

0

2 = 1

De este modo, tenemos que la velocidad se anula al primer segundo y a los cuatro segundos.

b. Recordemos que:

a(t) =

dv

dt =

d

dt(6t

2 − 30t + 24) = 12t − 30

Ahora sea t

0

el instante en que la aceleraci´on se anula, entonces a(t

0

) = 0

Ahora:

0 = a(t

0

) = 12t

0 − 30

As´ı tenemos que: t

0 =

30

12 =

5

2

Por lo tanto, la posici´on en este instante es:

x(t

0

) = x

5

2

= 2

5

2

3 − 15

5

2

2 + 24

5

2

+ 4 = 125

4 − 3

125

4 + 60 + 4 = −2

125

4 + 64 = −

125

2 +

128

2 =

3

2

De este modo, la posici´on de la part´ıcula cuando la aceleraci´on es cero es de 3

2 metros.

Adem´as la distancia total recorrida esta dada por:

distancia = |x(t

0

) − x(0)| = |

3

2 − 4| =

5

2

Finalmente la distancia total recorrida es: 5

5 0
3 years ago
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