For the maths boffins

If you can work out how far it's travelled in 6 mins it should be easy. I can't !

6 minutes 45 seconds. (I think).

If it's on the Jubilee Line, about three f**king hours.

Wheres Mart when you need him . ??

Are you sure OP's son read the question right? Six minutes to accelerate to 55 kph seems absurdly slow.

his sons only 7

Mar 25, 2020 -- 12:16PM, Ramruma wrote:

Are you sure OP's son read the question right? Six minutes to accelerate to 55 kph seems absurdly slow.

Not when a few members of this forum are on board.

I will guess 6mins 58 seconds

Howard, neither can I lol

as ram says...approx 2 min to get to 11mph..another 2 mins to reach 23mph..another 2 mins to reach 34mph..

maybe thats irrelevant and its only for the purpose of the calculation.

I was thinking, probably miles out but if you work it out per second 55 divided by 360, gives you 0.152 km per second.

then again I got a D for Maths O'Grade 

Is the answer "FISH" ?

I got 0 level maths, far too long ago now though...even the grandsons homework is hard lol

https://www.youtube.com/watch?v=s71atR_VHPs

A little something like what I wrote earlier!

For the first 6 minutes the train averaged 55/2 km/ph, (the average between 0 km/ph and 55 km/ph) and so would have travelled 2.75 kms in the 6 minutes. That leaves 7.25 kms to go at 55 km/ph which would take approx another 7.9 minutes. So the 10km journey will have taken 6 minutes plus 7.9 minutes making a total of 13.9 minutes.

Yeah, that's right. I forgot to divide the first half by 2 so had it travelling twice as far as it was meant to do in the first six minutes.

Exactly

Aye leave it to the experts, all makes sense now

Need more information here ?

What sort of snow is on the line please ?

Cheers uwgs.

Dealer, you sound like me lol

Is that correct? The train reached 55km/p/h accelerating uniformly, aren't you assuming that it will travel at 55km/p/h for the rest of the journey?
Can't recall velocity from donkeys years ago but i assumed the train would continue to accelerate.

Just a slight dent in uwg methodology
If the average was 55kph then it must have travelled faster at some point
The question doesnt quote the average though
GL

Surprised you don't know the answer Koi
Being red hot at point to point

average wasn't 55.....max was.....uwg is correct

he said averaged 55/2 which is 27.5 kmph

just out of interest , where was the train going ?

It takes 6 min to reach 55km/hr and covers 2.75km......leaving 7.25 km to go @55km/hr.
i.e. 7.91 mins...total time 13.91 mins..

Think of it like this...the end velocity is 55km/hr,....or approx 1 km /min.
So, to cover the whole 10 @1km/ min would take 10 mins...but its first got to reach the end speed which takes 6 min.
This takes up 2.75 km...leaving 7.25 @approx 1 km/min...

Maths is not in question
Assumption of top speed is
Call Stewart R for the defence
Old proverb once told  - Assume makes an ass of u and me

Was it an essential journey btw
& did they have any egg & cress sandwiches available

Stick to me times tables

11 minutes 26.49secs to 2dp

work in metres and seconds using

v = u + at
and
s=ut+at^2/2

starts from rest so u = 0

we know v (55kph at time t 6 minutes, hence derive a)
want t at s= 10k so plug in derived a and derive t

What's the voltage and how big is tthe flux capacitor

9.166666666666 mins  would be the answer I think they are looking for but it's prob completey wrong, not had a look at the replies yet,the lack of info in the question is typical of school **** now.

What age group is this for?

Oops

Cell 1	Cell 2
Cell 3	Cell 4

It only asks how long to travel 10km, not how long to travel the first 10km of a journey.Its a trick question imo.

v = u + at
and
s=ut+at^2/2

Well, those equations take me back. Probably A-level maths classes

s=1/2(u+v)t to find the distance traveled in first six minutes. Then 10-ans to give the rest of the distance inthe journey. This can then be divided by 55 multiplied by 60 to give the time for the rest of the journey. Add firstsix to this to give thetotal length of time.

^^,no,  when i was at school 50s/early 60s this was o level maths

s=1/2(u+v)t to find the distance traveled in first six minutes. Then 10-ans to give the rest of the distance inthe journey. This can then be divided by 55 multiplied by 60 to give the time for the rest of the journey. Add firstsix to this to give thetotal length of time

13.9..

Mar 25, 2020 -- 5:03PM, 1stpost wrote:

^^,no,  when i was at school 50s/early 60s this was o level maths

>When schools say exams aren't getting easier, ignore them (to a point definitely). For our a-level practice, we did previous year's o-level papers as that's how much the maths syllabus had changed.