How Do You Calculate Constant Deceleration to Prevent a Train Collision?

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To calculate the constant deceleration needed to prevent a train collision, the positions of both the locomotive and the passenger train must be set equal, accounting for their speeds and the initial separation of 0.42 miles. The locomotive travels at 18 mi/h while the passenger train moves at 100 mi/h, requiring the use of relative speed and time to determine the necessary deceleration. It is essential to convert all measurements to consistent units, preferably feet and seconds, for accurate calculations. The solution involves analyzing the area under speed vs. time graphs for both trains to ensure the distance covered by the decelerating train does not exceed that of the locomotive. The calculated deceleration to avoid collision is 3.26 ft/sec².
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Say a locomotive at a constant 18 mi/h comes onto a track (from some other track) that has a passenger train att 100 mi/h on it. At the instant the train is 0.42 miles away from the locomotive, it starts to decelerate. What is the resulting constant acceleration that is needed to just avoid a collision? I have 18t + 0.42 = 100t + 0.5a(t^2).

However, I can't solve this! I know there's something simple I haven't seen, so please help.

Thanks.
 
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OVB said:
Say a locomotive at a constant 18 mi/h comes onto a track (from some other track) that has a passenger train att 100 mi/h on it. At the instant the train is 0.42 miles away from the locomotive, it starts to decelerate. What is the resulting constant acceleration that is needed to just avoid a collision? I have 18t + 0.42 = 100t + 0.5a(t^2).

However, I can't solve this! I know there's something simple I haven't seen, so please help.
I assume the locomotive and train moving in the same direction. What is the relative speed and separation as a function of time? What is the condition for no collision?AM
 
Last edited:
Andrew Mason said:
I assume the locomotive and train moving in the same direction. What is the relative speed and separation as a function of time? What is the condition for no collision?


AM
Well, they are in the same direction, and the point is to set their position functions equal so that they are a centimeter (or something that negligible) away from touching)

The only info my problem gives is that one is at 18 miles an hour, the other at 100 mi/hr, and the separation as 0.42 miles. This is getting to me, because evidently I would have to solve for time AND acceleration.
 
Do I need calculus to solve the problem by any chance?By the way, the answer in my book is 3.26 ft/sec^2, but I don't know how to arrive at that...
 
Units! Units! Units!

ALWAYS use ft and seconds for units in kinematics. CONVERT everything before you even start the problem.
 
OVB said:
Do I need calculus to solve the problem by any chance?


By the way, the answer in my book is 3.26 ft/sec^2, but I don't know how to arrive at that...
Draw a graph of the speed vs. time of each. The area under each graph between two times is the distance covered in that time interval. So, do an expression for those areas and apply the condition (ie where the area under the train graph is less than the area under the locomotive graph).

AM
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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