Applied Maths question

Prowetod

This question I cant seem to get out, any help would be appreciated.

A train accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a period of time and then decelerates uniformly to rest. If the average speed for the whole journey is 5v/6, find what fraction of the whole distance is described at a constant speed.

PurpleFistMixer

Okay. Speed time graphs are always your friend for these kind of questions, helps to make things clear...

Anyway, this is what I got...

The average speed is just the total distance over the total time, so...
D/T = 5v/6

Now (I did this a slightly convoluted way as I thought it was the TIME, not distance you were looking for, so there's probably unnecessary steps in here, but I have dinner to eat and I'm not doing it again : p), separate your graph into 3 regions - the accelerating, the constant, and the decelerating.
Now, I assumed the acceleration and the deceleration were the same... it would have been trickier otherwise... normally that sort of thing is clarified in a question (or at least one hopes).

So, your D comes out as being (Ta)v + (Tc)v (where Ta is the time spent accelerating/decelerating, and Tc is the constant speed time). (Using the fact that the distance travelled is the area under the graph!)
However, you know T (total) is just 2Ta + Tc.

So, you put your new value for D over your new value for T and let that equal 5v/6... the vs cancel, and solving for Tc you get that it's 2T/3.

Which is when I realised that's not what the question was asking (silly me)...

So, anyway, the distance travelled in this time is just 2Tv/3 anyway, and you know your total distance is 5vT/6, so you put one over the other and arrive at... 4/5. Which is possibly the answer. ... If any of that makes sense, and if I didn't make any hideously glaring errors!

MathsManiac

I got the same answer as PFM, but without assuming the deceleration was equal to the acceleration. I believe the deceleration can't be determined from the information given, but that the answer is independent of the deceleration, which is why the assumption still led to the same answer. (Choosing a different deceleration would yield a different time spent at constant speed, but the answer would still be the same.)

Anyway, here's how I did it.

Speed-time graph is, as already stated, the key. The area is the distance travelled. Take d1, d2, and d3 for the distances travelled in the three chunks of the journey (i.e. the area of triangle, rectangle, triangle).

Now imagine the train (or another train) had done the whole journey at a constant speed in the same time. That is, the average speed. Put this on the diagram too. (See attachment.)
train.jpg
For these two areas to be equal the red, pink and purple bits have to equal the two blue bits.

The red triangle is one sixth as tall and one sixth as wide as d1, so its d1/36. Similarly, the purple triangle is d3/36. And the pink rectangle is d2/6.

The blue triangle on the left is 5/6 as wide and 5/6 as high as d1, so it's 25/36 times d1. Similarly, the one on the left is 25/36 times d3.

Equating and multiplying by 36 gives d1 + 6d2 + d3 = 25d1 + 25d3.
This leads to d2 = 4(d1+d3), so 5d2 = 4(d1+d2+d3). That is, d2 = (4/5)(d1+d2+d3)

straight_As

eoccork wrote: »

This question I cant seem to get out, any help would be appreciated.

A train accelerates uniformly from rest to a speed v m/s. It continues at this constant speed for a period of time and then decelerates uniformly to rest. If the average speed for the whole journey is 5v/6, find what fraction of the whole distance is described at a constant speed.

We did this in class last week. Here's how I did it.

Average speed for whole journey = 5v/6 => distance = 5vT/6 (T=total time)

Draw graph with periods of acceleration, constant v and deceleration.
Max velocity = v

=> vT would be the distance traveled with no accelerations or decelerations.

=> vT - 5vT/6 = distance traveled (combined) during acceleration and deceleration = vT/6.

=> Distance traveled at constant v = 5vT/6 - vT/6 = 4vT/6

=> Fraction of total d traveled at v = (4vT/6)/(5vT/6) = 4/5

If there's any mistakes, please point them out.

straight_As

This should explain my thinking better.

Sorry about the crappy graph.

MathsManiac

straight_As wrote: »

If there's any mistakes, please point them out.

No mistakes that I can see! That's a neat solution.

dr.pharmacy

Good Quation
answer next time

mardybumbum

How I miss that subject...........:rolleyes: