1. ## Physics poser

K, I have a question here from my daughter. Is this a place with a Physics major? is this an area for light-hearted fun, or can a question be answered like this one:

A train 350 m long is moving on a straight track with a speed of 83.4 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 18.0 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.

She needs a bit of explaining as to how to calculate the result. She has several posers like this, but thinks if just one were explained rationally to her she would understand.

Thanks everyone,
nannette

2. ## Re: Physics poser

Using a variation of the time, speed and distance formula, I came up with <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>24.852 </font color=yellow></span hi> seconds.

The formula is: speed = distance / time . I've attached my spreadsheet that walks you through the process but be forewarned that this is not my line of work! I would encourage you to wait for one of the lounge masterminds to chime in with a response before I'd share the theory with others...

I used the average of the two speeds that were given and I'm just not 100% sure this is the correct way. The formula may only work with a single constant speed?? I am ceratin that before you can come up with the correct solution, you'll have to break things down to simpler terms: meters instead of km, and seconds rather than hours.

I had fun with this...hope I'm right. <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

3. ## Re: Physics poser

<P ID="edit" class=small>(Edited by sdckapr on 06-Sep-03 20:41. OOPS. Corrected after reviewing Ricky's solution.
I used 834 km/h instead of 83.4! This answer (and Rick's) is correct.
)</P>Not a physics major but a chemist, but I have had calculus and physics in HS and college:
The 3 General Eqns to KNOW for ALL these types of problems:

1)Velocity(final) - Velocity(initial) = acceleration * time

2)Distance (final) - Distance(initial) = velocity(initial) + 1/2 * acceleration * time^2

3)[Velocity (final)]^2 = [Velocity (initial)]^2 + 2 * acceleration * [distance(final)-distance(initial)]

1 and 2 are sufficient to solve the puzzle. combine to eliminate the acc term, then solve for time.

Another way to do this knowing the general equation that the:

distance(traveled) = Velocity * time
If you have a CONSTANT acceleration, you can calculate the time to travel a distance using the AVERAGE velocity.

The answer I get is 24.9 sec. I leave the solution to your daughter. BOTH methods will yield the same value.

Steve

4. ## Re: Physics poser

Most all physics courses and texts approach the generic topic "straight line kinematics under constant acceleration" in much the same way. They derive a set of handy formulas from the straight-line graph of speed vs. time. These formulas (which I'm sure will appear in your daughter's text in some form) relate the quantities:
t time interval
vi initial speed
vf final speed
a acceleration
d distance travelled

All quantities relate to the particular time interval of interest, and all can be positive or negative (except for t). The key to solving such problems is to determine which of these quantities is being sought, and which one is not specified or readily accessible. This then points to which formula to use. In this case, a is not specified (other than verifying it is constant) and t is being sought. The appropriate formula will look something like:
d = (1/2)(vf + vi)t

It's rather more illuminating to look at a more "down to Earth" interpretation of these things, than blindly pulling the formula that fits out of the bag of tricks. The above is really just a statement that the distance travelled in the time interval is the average speed over that time interval, multiplied by the amount of time elapsed. In this case (and only for the case of constant acceleration) the average speed is the simple arithmetic mean of the speeds at the beginning and end of the time interval (because the speed varies linearly with time, because the acceleration is constant).

Common pitfalls:
<UL><LI>All of these handy formulas apply only for constant acceleration and motion along a straight line.
<LI>Work in consistent units - usually metres, seconds
<LI>Clearly define which direction is called positive, and which is negative. In this case, if the direction of motion of the train (and hence the speeds) are positive, then the acceleration will be (correctly) negative.[/list]I may have (somewhere) a little summary in Word of all this stuff I taught years back. It's geared away from the "find the formula" approach (all the textbooks do that anyway) and more towards the physical interpretation of the situation. I'll post if you're interested.

Alan

5. ## Re: Physics poser

I think some of what you've stated explains why I wasn't 100% sure of my answer. When I was putting the numbers to a graph, it occurred to me that if the aceleration (or deceleration) line had been anything other than "straight", the elapsed time would've been different. My only protection was the statement in the puzzle that mentioned "constant" speed.

6. ## Re: Physics poser

The question actually says "Assuming constant acceleration", which means that the gradient of the speed vs. time graph will be constant. This, in turn, means that the average speed over the time interval will be the simple arithmetic mean of the initial and final speeds. The more general equation, which always holds is
d = v(ave) x t

which is simply the definition of average speed over a time interval. For cases other than constant acceleration, this becomes rather more complex than the simple arithmetic mean of the two end values.

Alan

7. ## Re: Physics poser

Thank you so much everyone for your help. I am printing out all your responses so she may read your explanations, and I am sure she is ever so grateful. Again...wowsers...so many thanks
Nannette

8. ## Re: Physics poser

I dug up what I think was a draft of some handout content re: kinematics. It shows how the standard equations are derived from the straight-line Speed vs. Time graph, and how certain of the quantities relate to each other by simple geometry. We sometimes added these sorts of supplements to the more standard treatments, in an effort to demystify the often overwhelming barrage of equations & formulae that seemed to appear with each new physics topic. Hope it's of use.

Alan

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•