HELP: Need help solving a problem

I need help solving a problem. I suspect I am just missing a small
piece of the puzzle. But unless I get some help, I'm not going to
figure this out on my own.

The physics problem is this:

A train, 100m is length, accelerates uniformaly from rest. If the
front of the train passes a railway worker 150m down the track at a
speed of 25 m/s, what will be the speed of the last car as it passes
the worker.

The problem is simple uniform motion, and constant acceleration.

Known:

d1 = 0
d2 = 150m
d3 = 250m

v1 = 0
v2 = 25 m/s
v3 = ?


I understand the 25 m/s to be velocity - but I need to find the
acceleration and time for any of the uniform motion equations to work.
To date, I've trie this problem two or three different ways, and came
up with the same number of answers.

Can someone explain the proper way to solve this problem.

Any help would be appreciated..

Pat

"Pat" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...

We have to make an assumption here. We don't know what the acceleration is.
We *are* told that the worker is "150m down the track". We must assume that
the specified distance represents how far he is **from the front of the
train when it starts**, because *that* piece of information *will* allow us
to calculate the acceleration, which makes the problem solvable.

Then we have (did your book go over this one?):

v2^2 = v1^2 + 2*a*(d2-d1)

[We get this equation from

v2 = v1 + a*t

and

d2 = d1 + (1/2)*(v1 + v2)*t

by eliminating t.]

Can you take it from there, Pat?

Sometimes a problem is badly or incompletely stated, and we need to make an
assumption like this one to get to a solution. We *assume* that there IS a
solution because we were given the problem by someone who probably thinks
there is a solution. This is an introduction for you to the need for the
ability in physical science to give *precise* descriptions of objects,
processes, et cetera.

My experience is that the vast majority of disputes in science can be traced
to errors or inadequacies in communication. Watch this newsgroup for more
examples of this phenomenon.

HTH


Tom Davidson
Richmond, VA

Hi Tom,

Thank you for the very quick response...

1) I need to read up a bit more on eliminating "t" so I actually
understand how to do this on my own. "I've been away from school
too long."

2) Going with your equation: V2sqared = V1squared + 2 (a) (d2-d1)

I get

625 = 0 + 2 (a) (150m)

i.e 25 m/s squared is 625 - this is the known velocity as the front
of the train passes the railway worked at a point 150m down the track.

Then again. Am I looking a this problem wrong. I'm assuming I have
to find the acceleration for the first part (150m) before I calculate
the speed at 250m which "too me" is the total displacement when the
last car would pass the railway worker.

The second question is "do we solve the Eq the way it is" or "are we
maipulating the Eq by moving from rightside to the left side. If so,
what moves to the left side of the equation.?

From what numbers I plugged into the Eq above:.

If we manipulate the Eq is it:

625 - 150 = 2 (a) or 625 - 300 = 2 (a)


Comment: I blew through my "Grade 12" physics and trying to tackle
pre-university level "after being away from school for several years"
it is amazing how much I have forgotten. Unfortunately the text
book is so condensed, it's assuming I will remember every detail
and those are the bridges I need to rebuild.

Anyway, in the mean time I'll break open the book and start looking at
eliminating "t" etc, and manipulating the equations.

Thanks...

Pat



On Sun, 18 Apr 2004 03:35:13 -0400, "tadchem"
<[Only registered users see links. ]> wrote:

"Pat" <Pat@nospam.com> wrote in message
newsvo48053tsgmchcvt5d4taan2gq2k7brhf@4ax.com...

Solve one of the equations for t. This should give you an expression "t =
(something)". Then substitute the (something) you get fo thte "t" in the
*other* equation and then algebraically 'clean it up.'


So far, so good.


Correct. As you put it in the first post the train "accelerates uniformaly
from rest". This means that the acceleration is constant until we are told
otherwise (presumably after it has gotten up to speed somewhere down the
track).


You seem to have forgotten a little algebra. Step by step is goes like
this:

625 = 0 + 2 (a) (150m)

625 - 0 = 2 * a * 150 = 625

625 / 2 = a * 150 = 312.5

312.5 / 150 = 2.083

The units are meters per second squared.


Let me recommend a review of basic algebra first.


Tom Davidson
Richmond, VA

Tom, I agree - I definitely think a review of algebra is in order...

I'm going to work on this tonight/tommorow and see what happens.

Pat





On Sun, 18 Apr 2004 11:00:12 -0400, "tadchem"
<[Only registered users see links. ]> wrote: