Ive been trying to figure this out for two days now with no luck The engineer of a passenger train traveling 25.0 m/s sights a frieght train''s caboose 200m on the same track. The frieght train is traveling at a constant 15.0m/s in the same direction as the passenger train. The engineer applies the brakes and the passenger train begins to decelerate at a rate of -.100 m/s^2, while the freight train continues at its constant velocity. Consider the front of the passenger train as x=0. a). Will the cows nearby witness a collision? b). if so where will it take place? c). on a single graph sketch a x t graph of both the trains positions. This is the only problem out of 20 others I can''t seem to get. Here''s what I know so far: Frieght_V=15.0 Frieght_Xinitial=200 Pass_V=25.0 Pass_A=-.100 Pass_Xinitial=0 So I assume: Pass_Xfinal=0 + 25t + 1/2(-.100)t^2 Frieght_Xfinal=200 + 15t so I solved for unknown: t = (x-200)/15 then plugged into the other eq: Pass_Xfinal=25( (x-200)/15 ) -.05( (x-200)/15 )^2 That''s just an idea I think that''s how you begin to solve it. If someone could steer me in the right direction I would appreciate it. The assignment is due tommorrow and I only have to finish this one problem. (oh and if I don''t its gonna bug me all night)

A Little Hint:

The first part of the question is to determine whether the trains crash or not. the condition for the trains to crash is that the distance between them is 0. you should be able to find an equation giving the distance between the trains in terms of time(youve almost done this already). if it is solvable then they crash.

Well Im still confused

Would that mean that the equation I have right now, if simplified, may result in a root that gives 0 distance between the trains? Or does that mean if I set the two equations equal? (meaning they have 0 distance between them)

You have an equation giving the position of the freight train:
200 + 15t

You also have an equation giving the position of the passenger train:
25t - 0.05t²

so the distance between the trains is the position of the freight train minus the position of the passenger train. if they have crashed then the distance between them will be 0

Just re-read your post

yes by setting the two equations equal then you are saying the position of the passenger train equals the position of the freight train. ie they have crashed

why didnt my superscript work?

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Ah so then a simple quadratic equation can be set up.

.05t^2 - 10t + 200 = 0

That gives roots of 177.5 and -22.5 m
I assume they are moving in the positive direction so 177.5 is the answer. So...

t=(177.5-200)/15 = collision occured @ -1.5 secs? That doesn''t seem right. :/

the quadratic is for t. it gives you the time of the collision.
so gives you roots of t = 177.5s and -22.5s as we started at 0s then the collision would have occured at 177.5s.

think about what the equation means

Hi

seems you got your variables messed up

Let v1 be the passenger trains velocity
v2 the other trains velocity
a is -0.1m/s^2
x0 is the 200m

equation for passenger train:
x = t*v1 + t^2 * a/2
freight train:
x = x0 + v2*t

equal them:

t*v1 + t^2 * a/2 = x0 + v2*t

reorder:

a/2 * t^2 + (v1-v2)*t - x0 = 0

solve for t with quadratic formula:

t = ( (v2-v1) +/- sqrt((v1-v2)^2 + 2*a*x0)) ) / a

fill in vars:

t = ( -10 +/- sqrt(60) ) / -0.1

t = 177,459...

now insert t into one of the train equations, gives same result with both:

for the freighter:

x = x0 + v*t = 200 + 15 * 177,46 = 2826

but you will propably do it with higher precicion

---

Runicsoft -- home of my open source Function Parser and more

The collision actually occurs at t = 22.5s at a distance of 538.1m the roots to the quadratic are both posotive

at t = 177.5 is when the passenger train has slowed down so much the the freight catches it.