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A passenger train is traveling at 29 m/s when the engineer sees a freight train 360 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 6.0 m/s. (a) If the reaction time of the engineer is 0.40 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided? (b) If the engineer’s reaction time is 0.80 s and the train loses speed at the minimum rate described in Part (a), at what rate is the passenger train approaching the freight train when the two collide? (c) For both reaction times, how far will the passenger train have traveled in the time between the sighting of the freight train and the collision?

Respuesta :

Answer:

Part a)

[tex]a = -0.75 m/s^2[/tex]

Part b)

[tex]a = -0.77 m/s^2[/tex]

Part c)

[tex]d_1 = 532.74 m[/tex]

[tex]d_2 = 530.3 m[/tex]

Explanation:

Part a)

As we know that the relative speed of the two trains is given as

[tex]v_{rel} = 29 - 6 = 23 m/s[/tex]

since the reaction time is 0.40 s

so the distance between two trains will reduce by

[tex]d = v_{rel} t[/tex]

[tex]d = 23(0.40) = 9.2 m[/tex]

now we have the distance between them

[tex]x = 360 - 9.2 = 350.8 m[/tex]

now the distance between them will become zero if the deceleration is "a"

now we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 23^2 = 2(a)(350.8)[/tex]

[tex]a = -0.75 m/s^2[/tex]

Part b)

If the reaction time is t = 0.80 s

so the distance between the two trains will be

[tex]x = 360 - (23 \times 0.80)[/tex]

[tex]x = 341.6 m[/tex]

now we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 23^2 = 2(a)(341.6)[/tex]

[tex]a = -0.77 m/s^2[/tex]

Part c)

Distance moved by the train in first case

[tex]x = v t + \frac{1}{2}at^2[/tex]

[tex]350.8 = 23 t - \frac{1}{2}(0.75) t^2[/tex]

t = 28.4 s

now the distance

[tex]d_1 = 29(0.40) + 29(28.4) + \frac{1}{2}(-0.75)(28.4)^2[/tex]

[tex]d_1 = 532.74 m[/tex]

When reaction time is 0.8 s

[tex]x = v t + \frac{1}{2}at^2[/tex]

[tex]341.6 = 23 t - \frac{1}{2}(0.77) t^2[/tex]

t = 27.6 s

now the distance

[tex]d_2 = 29(0.80) + 29(27.6) + \frac{1}{2}(-0.77)(27.6)^2[/tex]

[tex]d_2 = 530.3 m[/tex]

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