train problem

[Poetess]

ok, heres a physics problem that i just got, and ive spent like 40 minutes and i havent figured it out yet...

when a high speen passenger train traveling at 161km/hr rounds a bend, the ingineer is shocked to see that a locclmotive has entered onto the track from a siding and is 676m ahead. the locomotive is going 29km/hr. the engineer of the high speed train immediately applies the brakes. what must the magnitude of the constant deceleration be if a collision is to be just avoided?
-i was never any good at physics

------------------
Age is opportunity no less
Than youth itself, though in another dress,
And as the evening twilight fades away
The sky is filled with stars, invisible by day.
-- Henry Wadsworth Longfellow


[This message has been edited by Poetess (edited 02-09-2002 05:33 PM).]

[Poetess]

nevermind...just figured it out, with some equations i had...but go ahead and answer if you want.

[Bicho the Inhaler]

...about 300,413.793 km/h^2?

[ctrlaltdel]

hmm, guess a few people will spill their drinks...

[Poetess]

no, i dont think so bicho.

------------------
Age is opportunity no less
Than youth itself, though in another dress,
And as the evening twilight fades away
The sky is filled with stars, invisible by day.
-- Henry Wadsworth Longfellow

[ralphmerridew]

The way I see it, a collision is unavoidable as the locomotive isn't braking.

[Poetess]

hehe, very funny , the locomotive is going the same direction as the train of course .

------------------
Age is opportunity no less
Than youth itself, though in another dress,
And as the evening twilight fades away
The sky is filled with stars, invisible by day.
-- Henry Wadsworth Longfellow

[Bicho the Inhaler]

Criminy...how about 12,887.6 km/h^2?

[Encarta]

1 2 3 4 5 6 7 8 9 10, am I getting closer

[tigg]

yes. keep going.

[Quailman]

Is that right, Bicho? Can you use 132km/hr as the relative speed that must be overcome before the train travels 676 m.? If that's the case, I get a maximum 36.87 seconds to slow to 29 km/hr. and a decelleration rate of 9,275.148 km/hr^2. How does one solve this problem?

[Chuck]

Computer simulation.

[Bicho the Inhaler]

Quailman, I'm not too sure about it; it's been many months since I've done any physics.

I seem to remember this equation for constant acceleration (which was assumed here): V^2 = (V0)^2 + 2AX, where V is the final velocity (0 km/h), V0 is the initial velocity (132 km/h), A is the (constant) acceleration, and X is the distance travelled (0.676 km), which made sense to use because it doesn't involve solving for time. When I solve that for A, I get -12,887.6 km/h^2.

I'll see if I can derive that from more basic equations:

We can always find distance travelled by multiplying average velocity by time, and for constant acceleration, the average velocity is simply (V0 + V)/2, and by definition V = V0 + A*T (where T is time), so for distance:

X = (V0 + V0 + AT)/2 * T = 1/2*AT^2 + (V0)T.

Also, V = V0 + AT, so V^2 = (V0)^2 + A^2T^2 + 2(V0)AT, and
A^2T^2 + 2(V0)AT = 2A(1/2AT^2 + (V0)T) = 2AX, so V^2 = (V0)^2 + 2AX.

***
Alternatively, if we can assume kinetic energy = 1/2MV^2 and that work = force * distance and force = MA, we know that energy will be conserved and we can set the two equal to each other thus:

1/2MV^2 = MAX, so cancel out M and...
V^2/2 = AX, which is what we had.

[Icarus]

Quailman - my immediate thoughts are, no, you cannot use 132 km/hr as the speed to overcome. The high speed train is constantly slowing down, yet the speed of the locomotive remains the same. If the question stated was "how long until a collision if niether train applies their brakes ?", then you could use 132 km/hr.

[Zarriar]

Hmmm.. probably wrong but anyway.

I have that the fast train has to deccelerate at -0.48 ms^-2


Just realised that I screwed up by a factor of 2



[This message has been edited by Zarriar (edited 02-11-2002 09:01 PM).]

[Lucky Wizard]

I get 12,266 km/hr^2, which is equal to 0.94641 m/s^2. I got it by solving this system of equations for x, a, and t:

code:

---

x = (1/2)at^2 + 161t





x = 29t + 0.676





0 = at + 161

---

[DJC]

I can see Lucky Wizard's first two equations. Should the third be 29=at+161?

Of course, all this does not allow for a realistic reaction time, which will cut into the 'head start'.

[tigg]

I’m pretty confident you can use relative velocities here. The acceleration would be the same regardless of your frame of reference. So you can just choose the locomotive as a point with 0 velocity and 0 acceleration. Then the fast train is coming at it 132km/h, and has 676m to stop.
Also, Bicho’s v^2 = 2ax is right as well.
I like working in m/s, so…
v = 132km/hr * 1000m/km * hr/3600sec = 36.66 m/s.
x = 676m, so a = .9944m/s^2.
Gravity is 9.8 m/s^2, so this is about 1/10 g.
I’ve been on roller coasters worse than that.

[Zarriar]

I got the same as you tigg, eventually! You must have rounded off more precisely than me however.