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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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A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).
just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles (\vec r_{cm}), measured in meters, is defined by this weighted average:

\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}} (1)

Where:

m_{i} - Mass of the i-th particle, measured in kilograms.

\vec r_{i} - Location of the i-th particle with respect to origin, measured in meters.

If we know that \vec r_{cm} = (-0.500\,m,-0.700\,m), m_{1} = 1\,kg, \vec r_{1} = (-1.20\,m, 0.500\,m), m_{2} = 4.50\,kg, \vec r_{2} = (0.600\,m, -0.750\,m) and m_{3} = 4\,kg, then the coordinates of the third particle are:

(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}

(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})

(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)

(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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