Find Min Distance for Train to Avoid Collision

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In summary: You don't know B. With your B= 130.0m^2/s^3, t= 0.414 seconds. That's the time it would take to increase from 11.5 m/s to 23.5 m/s with a= B/v. However, if you want to find the time it takes for this car to go from 11.5 m/s to 23.5 m/s accelerating at constant power, you need to use the work-energy theorem. Work= change in kinetic energy. The initial kinetic energy is (1/2)mv^2= (1/2)(1000)(11.5)^2= 132250 J.
  • #1
ffrpg
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I know I've asked quite a few questions, but I really am making an effort to get as far as I can. The engineer of a train moving along a level track with a velocity of 42.0 m/s sights a freight train at a distance of d ahead of him on the same track moving in the same direction with a velocity of 18.0 m/s. He applies the brakes, giving his train a constant acceleration of -1.4 m/s^2. What is the minium distance d such that there is no collision?


Here's what I came up with. I used the formula d/delta x = 42.0t+1/2(-1.4)t^2. I derived the formula giving me 42+(-1.4)=18. Solved for t getting 17.143. Plugged t back into the original formula I used and ended up getting 514.288. That isn't the correct distance..



Here's another question..
Let's say that you are driving a car that accelerates according to a=B/v where B= 130.0m^2/s^3 is a parameter that is related to the ratio of your car's power to its weight, and c is your car's speed, in m/s. Assume that you are initially traveling with a speed of 11.5 m/s. At t=0 you step on the gas pedal. The car performs a constant-power acceleration until you reach a speed of 23.5 m/s. What is the time interval needed to make this change of speed.


I solved for a by plugging in the values given for B and v, so a=130.0m^2/s^3/11.5 m/s. I then got a= 11.3m/s^2. From there I went to 11.5+11.3t=23.5. Solved for t and got 1.062 s. 1.062s isn't the correct answer. Was my approach to this problem incorrect?
 
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  • #2
Your first question is equivalent with the following: how far does a stone go up (what height) if one throws it upwards with (42-18)m/s and the gravitational pull is g=1.4m/s2 (about the same as on the moon).
For the second one I have another advice for you: start thinking the problems "in letters" first and then plug in the numbers. You have a*v=B=constant (as the power remains constant at those 175 horse power it has). you also have a=dv/dt. So I now see a diiferential equation with separable variables v and t. Do you know how to solve this one?
 
  • #3
Nope I still don't know how to solve it (the 2nd question). How did you figure out the car had 175 horse power? I don't think it matters. Are you saying I should integrate a=dv/dt?
 
  • #4
The engineer of a train moving along a level track with a velocity of 42.0 m/s sights a freight train at a distance of d ahead of him on the same track moving in the same direction with a velocity of 18.0 m/s. He applies the brakes, giving his train a constant acceleration of -1.4 m/s^2. What is the minium distance d such that there is no collision?


Here's what I came up with. I used the formula d/delta x = 42.0t+1/2(-1.4)t^2. I derived the formula giving me 42+(-1.4)=18. Solved for t getting 17.143. Plugged t back into the original formula I used and ended up getting 514.288. That isn't the correct distance..

Looks to me like you have done this correctly. you have x= 42.0t+ (1/2)(-1.4)t^2 (that's the distance the train goes in time t. I don't understand the "d/delta" part.) Differentiating (which in my opinion is better than "deriving") gives velocity and, of course, the train can avoid a collision by slowing to 18 m/s: dx/dt= 42.0- 1.4t= 18 (you left the t out in your formula. I hope that was a typo) which gives -1.4t= 18-42= -24 so t= -24/-1.4= 17.1 seconds precisely what you got. Now put that back into the formula for x: 42.0(17.1)+(1/2)(-1.4)(17.1)^2= 718.2-0.7(292.41)= 718.2-204.7= 513.5 m. Now, why do you say that is not the correct distance?

Let's say that you are driving a car that accelerates according to a=B/v where B= 130.0m^2/s^3 is a parameter that is related to the ratio of your car's power to its weight, and c is your car's speed, in m/s. Assume that you are initially traveling with a speed of 11.5 m/s. At t=0 you step on the gas pedal. The car performs a constant-power acceleration until you reach a speed of 23.5 m/s. What is the time interval needed to make this change of speed.
a= B/v is the same as dv/dt= Bv or dv/v= Bdt. Integrating both sides, ln|v|= Bt+ Const or v(t)= e^(Bt+ Const)= e^(Bt)e^Const= C e^(Bt) where C= e^Const. Since your initial speed is 11.5 m/s,
v(0)= C e^0= C= 11.5 so v(t)= 11.5 e^(Bt). Your car reaches 21.5 m/s when v(t)= 11.5 e^(Bt)= 21.5 or e^(Bt)= 21.5/11.5= 1.87. Now take the natural logarithm of both sides: Bt= ln(1.87). Since you know B you can now solve for t.
 
  • #5
Originally posted by ffrpg
Nope I still don't know how to solve it (the 2nd question). How did you figure out the car had 175 horse power? I don't think it matters. Are you saying I should integrate a=dv/dt?
I figured the car weighs about 1000kg so power=B*1000Kg.
I'm saying you should do the following:
v*dv=B*dt
integrate: ∫11.523.5 dv*v= ∫0t' dt'*B
giving: (1/2)v2|11.523.5 = (23.52-11.52)/2 = B*t

HallsofIvy: it's a=B/v, not a=Bv. Man, this would be easy if we could use the latex &over
 
  • #6
Oops! I switch the formula in mid-sentence!

You say, first, "you are driving a car that accelerates according to a=B/v" and then later say:"The car performs a constant-power acceleration" and seem to be assuming that means a constant acceleration. You calculate B/v for v= 11.4 m/s and then use that constant acceleration. I am inclined to interpret "constant power acceleration" as meaning that the acceleration follows the law you gave: dv/dt= B/v.

dv/dt= B/v gives vdv= Bdt so (1/2)v^2= Bt+ C. Your initial speed is 11.4 m/s so (1/2)(11.4)^2= C or C= 64.98. Your speed at any time t is given by v^2= 2Bt+ 129.96. You want to increase to 23.5 m/s so you need to solve 2Bt+ 129.96= 23.5^2= 16889.
 

What is the "Find Min Distance for Train to Avoid Collision" problem?

The "Find Min Distance for Train to Avoid Collision" problem is a mathematical and scientific problem that aims to determine the minimum distance that two trains traveling at different speeds must maintain in order to avoid a collision.

Why is it important to find the minimum distance for trains to avoid collision?

This problem is important because it helps ensure the safety of train passengers and the efficient operation of railway systems. By determining the minimum distance for trains to avoid collision, railway companies can implement appropriate safety measures and protocols to prevent accidents and disruptions.

What are the factors that affect the minimum distance for trains to avoid collision?

The minimum distance for trains to avoid collision is affected by various factors such as the speed of the trains, the braking capabilities of the trains, the reaction time of the train operators, and the length of the trains. Other factors such as weather conditions and track conditions can also play a role.

How is the minimum distance for trains to avoid collision calculated?

The minimum distance for trains to avoid collision is calculated using mathematical equations that take into account the factors mentioned above. These equations consider the speed, distance, and reaction time of the trains to determine the minimum safe distance that they must maintain.

How can the minimum distance for trains to avoid collision be applied in real-life situations?

The minimum distance for trains to avoid collision can be applied in real-life situations by railway companies to establish safe operating protocols and to design and maintain railway infrastructure. It is also used in the development of train signaling systems and the planning of train schedules to prevent collisions and ensure the smooth flow of train traffic.

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