Deceleration necessary to prevent a train collision

In summary, the problem involves an engineer of a passenger train who sights a freight train ahead on the same track. The passenger train is traveling at a certain speed, while the freight train is traveling at a different speed in the same direction. To avoid a collision, the engineer applies a constant deceleration to the passenger train while the freight train continues with constant speed. The question asks for the range of decelerations that will avoid a collision, with given values for the speeds and distance between the trains. An attempt at solving the problem by substituting for time and using equations for both trains was found to be incorrect.
  • #1
rezal
5
0

Homework Statement


The engineer of a passenger train traveling at vP sights a freight train whose caboose is distance D ahead on the same track. The freight train is traveling at vF in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant deceleration of a, while the freight train continues with constant speed.
For what range of decelerations a will the train collision be avoided?
Givens: vP,vF,D,

Homework Equations


Passenger Train: D=vP-.5at^2
Freight Train: x(t)=vFt
vP=vF+at

The Attempt at a Solution


I [/B]planned on solving this by substituting
t=(vP-vF)/a
into the equation for the passenger train
D= vP((vP-vF)/a) - .5a((vP-vF)/a)^2
and then solving for a but I was told that this is incorrect?
 
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  • #2
rezal said:

Homework Statement


The engineer of a passenger train traveling at vP sights a freight train whose caboose is distance D ahead on the same track. The freight train is traveling at vF in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant deceleration of a, while the freight train continues with constant speed.
For what range of decelerations a will the train collision be avoided?
Givens: vP,vF,D,

Homework Equations


Passenger Train: D=vP-.5at^2
Freight Train: x(t)=vFt
vP=vF+at

The Attempt at a Solution


I [/B]planned on solving this by substituting
t=(vP-vF)/a
into the equation for the passenger train
D= vP((vP-vF)/a) - .5a((vP-vF)/a)^2
and then solving for a but I was told that this is incorrect?
It is incorrect, but who told you that, and did they give you a reason?
 
  • #3
My TA did but he didn't really give any particular reason as to why.
 

1. What is the concept of "deceleration necessary to prevent a train collision"?

The deceleration necessary to prevent a train collision refers to the rate at which a train must slow down in order to avoid colliding with another train or obstacle on the tracks.

2. How is the deceleration necessary to prevent a train collision calculated?

The deceleration necessary is calculated by taking into account the speed of the train, the distance to the obstacle or other train, and the reaction time of the train operator.

3. What factors can affect the deceleration necessary to prevent a train collision?

The deceleration necessary can be affected by various factors such as the weight and length of the train, the condition of the tracks, the weather conditions, and the braking system of the train.

4. How can the deceleration necessary to prevent a train collision be achieved?

The deceleration necessary can be achieved by implementing proper safety measures such as maintaining safe speeds, regular maintenance of trains and tracks, and using advanced braking systems. It is also important for train operators to be alert and follow safety protocols.

5. What are the consequences of not achieving the necessary deceleration to prevent a train collision?

Not achieving the necessary deceleration to prevent a train collision can have serious consequences, including damage to the trains and tracks, injuries or fatalities to passengers and crew, and disruptions to train schedules. It can also result in costly legal and financial repercussions for the railway company.

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