Two trains headed toward each other

[goaliejoe35]

Homework Statement

As two trains move along a track, their conductors suddenly notice that they are headed toward each other. Figure 2-27
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c02/pict_2_27.gif

gives their velocities v as functions of time t as the conductors slow the trains. The figure's vertical scaling is set by vs = 43.0 m/s. The slowing processes begin when the trains are 204 m apart. What is their separation when both trains have stopped?

The attempt at a solution

I tried this problem and got it wrong. I was able to come up with d = 107.5 m for Train 1 but I think I am missing something when I try to calculate d for Train 2.

The part of this question that confuses me the most if when it says "The figure's vertical scaling is set by vs = 43.0 m/s." I think I don't really understand how to read the graph correctly.

Help is appreciated

 

[Kurdt]

Ok well, if 4 blocks up the side is 43m/s then what are three blocks going to represent for the velocity?

 

[Doc Al]

I tried this problem and got it wrong. I was able to come up with d = 107.5 m for Train 1 but I think I am missing something when I try to calculate d for Train 2.

You found the distance traveled by Train 1 before it stops. Do the same thing for Train 2.

The part of this question that confuses me the most if when it says "The figure's vertical scaling is set by vs = 43.0 m/s." I think I don't really understand how to read the graph correctly.

I take that to mean that V_s = 43 m/s--which means that 4 vertical boxes = 43 m/s. So what's the initial velocity of Train 2?

 

[goaliejoe35]

Well then is this right?


if velocity equals -33 m/s then...

Train 2

Data:
v = -33 m/s
v2 = 0 m/s
t = 4s
a = ?
...then I used this equation...
v2 = v + a2t
a2 = (v2 - v) / t

= (0m/s - (-33 m/s)) / 4s

a = 8.25 m/s^2

v2^2 = v^2 + 2ad
d = (v2^2 - v^2) / 2a

= ((0m/s)2 - (-33 m/s)2) / (2(8.25m/s^2)

= -66m <--distance traveled by train 2

 

[Kurdt]

It doesn't equal - 33m/s exactly. There are a couple of easier ways to calculate the distance traveled. Firstly one can calculate the area under the curve to the x-axis, or you can use the average speed formula which amounts to the same thing. There's nothing wrong with the method you've used, just a bit long.

 

[goaliejoe35]

I don't quite get what you are saying by I can use the average speed formula? Isn't that formula S(avg)=(total distance traveled)/(time elapsed)? How does that relate?

 

[Kurdt]

I don't quite get what you are saying by I can use the average speed formula? Isn't that formula S(avg)=(total distance traveled)/(time elapsed)? How does that relate?

No you're right, ignore that its a load of rubbish I was thinking of something else. Any way my main point was 33m/s is not quite the right velocity.

 

[goaliejoe35]

Ok so if 33 m/s isn't quite right how much is it off by and how else would I do this? I tried getting to the final answer using 33 m/s and my final answer was wrong so I am assuming that 33 m/s is off enough to cause me to get the problem incorrect.

 

[Kurdt]

What have you done to get 33m/s as the speed? Remember that 4 divisions on the y-axis is equivalent to 43m/s. What does that make 1 division equivalent to? So what velocity does 3 divisions correspond to?

 

[goaliejoe35]

Oh man...now i think i got it! 32.25 m/s ?

 

[Kurdt]

Oh man...now i think i got it! 32.25 m/s ?

Sounds good.

 

[rootX]

Ok so if 33 m/s isn't quite right how much is it off by and how else would I do this? I tried getting to the final answer using 33 m/s and my final answer was wrong so I am assuming that 33 m/s is off enough to cause me to get the problem incorrect.

If the answer is 32 m

try using areas method as mentioned before..


edit: seems like you are doing something else wrong too.. because there isn't much different between 33 and 32.5 (or maybe ...)