Physics Constant Acceleration Question?

Question

When a high-speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D= 676 m ahead. The locomotive is moving at 29.0 km/h. The engineer of the high-speed train immediately applies the brakes. What must be the magnitude of the resulting deceleration if a collision is to be just avoided?

Answer 1

So, apples to apples, the chasing deceleration is about 1 m/sec^2 and the oncoming case is a tad over 2 m/sec^2-- or a tenth and two tenths of a G respectively.

We have no data on the masses of the trains, and thus the energies involved...

Anybody got a clue about the performance of real-life trains, whether such decelerations are possible, and whether the the train drivers in the oncoming case would feel anything like that pilot who took off from the short runway last week? ;)

...Answering (sorta) my own question-within-a-question...

If "Anna" is to be believed, the French TGV train goes from 340 clicks (94.4 m/sec) to zero in 1200 meters. That works out to about 0.38 G-- or almost twice the deceleration of the oncoming case.

Answer 2

I also took it to mean that the trains were approaching each other but I see and agree with Ronin's viewpoint of it.

Assuming it to be that way:
Look at it from the point of view of the back bumper of the slow locomotive. Then the high-speed train is approaching at 132 km/h and must stop (relative to the locomotive) in d=676 meters.

Using
V^2 = Vo^2 +2ad where V=0, Vo=132 km/h and d=676 m,
I get 0.9944 m/s^2

Answer 3

The trains are traveling toward each other at a velocity of 190 km/h. They have 676 meters (.676 km) to get to 0 km/hr.
Using the appropriate formula, vt^2 = vi^2 + 2 as, we get
0^2 = 190^2 + 2 a .676
-36100 = 1.352 a

a, the deceleration, is about 26700 km/hr/hr

For some reason, we rarely see accelerations given in these units, but it works. If you want to convert to meters per second per second, you may do so.

Answer 4

This is how I did the question:

161km/h ------------- 29km/h
------------> {676m} ------------->

In a second, the train would travel 161000m / 3600sec = 44.72m.
In a second, the locomotive would travel 29000m / 3600sec = 8.06m.

So the distance between them would become 676 - 44.72 + 8.06 = 639.34m, which is 36.66m closer than they were before. Since they are travelling at constant velocity (I am assuming that), their distance would get 36.66m closer every second. That leaves us with 676m/36.66m = 18.44sec to stop the train before collision.

So, now that you know the velocity and time needed, you can find the deceleration simply by dividing the velocity by time.

The velocity is 161km/h and the time is 18.44sec, which is equal to 0.00512 hour, so the deceleration is 161/0.00512 = 31, 435km/h/h.

Answer 5

Collision cant be avoided because the driver of the locomotive(29.0 km/h) is not applying brake.

Answer 6

i think which you will have 2 varieties of graphs, the 1st form of graph is parabolic. occasion I drop an apple off of a tree and till it hits the floor that's in consistent acceleration, and its speed is increasing. interior the 2nd case, speed could be fastened yet you have consistent acceleration. occasion a satellite tv for pc orbiting the earth. (anticipate the earth is the middle of the universe because of the fact in case you do no longer then you definitely might desire to account the orbit of the earth around the sunlight, the sunlight around the middle of the milklyway galaxy, and on and on. i'm hoping this facilitates

Answer 7

I don't think he meant towards each other. SInce they are only 676 m apart, I assume he meant that they are going inthe same direction.