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anonymous

  • one year ago

KINEMATICS QUESTION -- ATTEMPT IF YOU DARE A passenger train is traveling at 30 m/s when the engineer sees a freight train 384 m ahead of his train traveling in the same direction on the same track. The freight train is moving at a speed of 6.1 m/s. (b) If the engineer's reaction time is 0.87 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? Answer from part a: 0.7696 I attempted part C too, but I'd like to wait for some help on B before I post my current work on C, if that's alright. Thank you!

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  1. anonymous
    • one year ago
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    What I have so far is: 30(t+.87) + .5(.7696)(t)^2 = 5.1t + 384 .7696 = (vf - 30)/(t) (system of equations) t = 25.113 or 37.09 and vf = 10.7051 or 1.49488 Why are there two answers? I know that there is a quadratic so it may have 2 roots but what do those two answers mean in terms of the context of the problem?

  2. anonymous
    • one year ago
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    |dw:1439680155506:dw| Not really sure if that helps but here's a picture

  3. welshfella
    • one year ago
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    i dont understand the question. What is the minimum rate of losing speed as described in part a?

  4. mathmate
    • one year ago
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    @bandicoot12 For a start: Even though you have attempted part (a), since the part (b) question refers to information given in part (a), you need to post the question for part (a) as well.

  5. anonymous
    • one year ago
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    If the reaction time of the engineer is 0.41 s, what is the minimum (constant) rate at which the passenger train must lose speed if a collision is to be avoided?

  6. anonymous
    • one year ago
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    30(t+.41) + .5(.a)(t)^2 = 6.1t + 384; rearranged it to -371.7+23.9t+.5(a)(t^2) = 0 23.9 + at = 0 (took the derivative to find the minimum acceleration)

  7. mathmate
    • one year ago
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    back now! sorry, meant to sent afk note.

  8. mathmate
    • one year ago
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    Well, part (a) goes like this: 1. relative speed between the two trains = s = 30-6.1=23.9 m/s 2. relative distance between two trains = 384 m. at start 3. relative distance between two trains after engineer applied brakes = 384-0.41s = 374.201 m 4. use kinematics equation to find a, acceleration: (v^2-u^2)=2ad => a=(v^2-u^2)/2d =(0^2-s^2)/(2*374.2001)=-0.76323954 m/s^2 5. time lapse before collision t=(0-s)/a=-374.201/-0.76323954=31.3138912 s 6. check total distance covered by faster train: D1=0.41*30+30t+(1/2)at^2= 577.515736 m 7. check distance travelled by freight train + 384 m D2=6.1*(t+0.41)+384=577.515736 m. 8. Since distances travelled match, we conclude a=-0.76323954 m/s^2

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