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Maru [420]
3 years ago
11

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

mueve a 5.00 m/s. Sue aplica los frenos, pero solo puede acelerar a -2.00 m/s2 porque el camino esta húmedo. ¿Habrá una colisión? Establezca como llega a su respuesta. Si es si, determine cuán lejos en el túnel y en qué tiempo ocurre la colisión. Si es no, determine la distancia de acercamiento mas próxima entre el automóvil de Sue y la camioneta.
Physics
1 answer:
Romashka [77]3 years ago
0 0

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}

Camioneta lenta

x_{C} = x_{C,o} +v_{C}\cdot t

Donde:

x_{S,o}, x_{C,o} - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

v_{S,o} - Velocidad inicial de Speedy Sue, medida en metros por segundo.

v_{S}, v_{C} - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

a_{S} - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

t - Tiempo, medido en segundos.

Si conocemos que x_{S} = x_{C}, x_{S,o} = 0\,m, x_{C,o} = 155\,m, v_{S,o} = 30\,\frac{m}{s}, v_{C} =-5\,\frac{m}{s} y a_{S} = -2\,\frac{m}{s^{2}}, encontramos la siguiente función cuadrática:

155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}

-t^{2}+35\cdot t-155 = 0 (Ec. 1)

Las raíces de esta función son:

t_{1}\approx 29.798\,s, t_{2} \approx 5.201\,s

La colisión ocurriría en la raíz positiva más pequeña, es decir:

t \approx 5.201\,s

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: (v_{C,o} = -5\,\frac{m}{s},  x_{C,o} = 155\,m, t \approx 5.201\,s)

x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)

x_{C} = 128.995\,m

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

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Replacing the values given,

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Answer:

5.9 cm

Explanation:

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Spring constant (K) =.?

F = Ke

15 = K × 3

Divide both side by 3

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Therefore, the spring constant (K) is 5 N/cm.

Now, we shall determine how far the spring will stretch when a 3 kg candy is placed on the spring balance.

This can be obtained as follow:

First, we shall determine the weight of the object.

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Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

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Finally, we shall determine how far the spring will stretch as follow:

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Therefore, the spring will stretch 5.9 cm when a 3 kg candy is placed on the spring balance.

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