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Bad White [126]
3 years ago
12

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

If the train's speed is 75 km/h, how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2–36. What are the results if the car and train are traveling in opposite directions?
Physics
1 answer:
SOVA2 [1]3 years ago
5 0

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be v_{AT}=v_A-v_T (<em>the train must see the car advancing at a lower speed</em>), where v_A is the speed of the automobile and v_T the speed of the train.

So we have v_{AT}=(95km/h)-(75Km/h)=20Km/h.

So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s

And in that time the car would have traveled (<em>relative to the ground</em>):

d=v_At=(95Km/h)(0.065h)=6.175Km

If they are traveling in opposite directions, <u>we have to do all the same</u> but using v_{AT}=v_A+v_T (<em>the train must see the car advancing at a faster speed</em>), so repeating the process:

v_{AT}=(95km/h)+(75Km/h)=170Km/h

t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s

d=v_At=(95Km/h)(0.00765h)=0.73Km

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To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

y = \frac{1}{2} at^2+v_0t+y_0

Where,

a = acceleration

t = time

v_0 = Initial velocity

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In addition to this we know that speed, speed is the change of position in relation to time. So

v = \frac{x}{t}

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With the data we have we can find the time as well

v = \frac{x}{t}

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With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

y = \frac{1}{2} at^2+v_0t+y_0

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6 0
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Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos: 

Altura o distancia recorrida: 40 m
Vo: 6 m/s 
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes. 

Entonces tenemos que:

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Sustituyendo tenemos que:

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Que tengas un buen día!
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