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i was wondering if you could help me??

i've bin stuck on a question for atleast 40 mins now and im no closer to finding the answer.

Question quoted from AQA M1 text book:

Telegraph poles, 40 m apart stand alongside a straight railway line. The times taken for a locomotoive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, repectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

i've bin stuck on a question for atleast 40 mins now and im no closer to finding the answer.

Question quoted from AQA M1 text book:

Telegraph poles, 40 m apart stand alongside a straight railway line. The times taken for a locomotoive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, repectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

Anujkhamar

i was wondering if you could help me??

i've bin stuck on a question for atleast 40 mins now and im no closer to finding the answer.

Question quoted from AQA M1 text book:

Telegraph poles, 40 m apart stand alongside a straight railway line. The times taken for a locomotoive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, repectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

i've bin stuck on a question for atleast 40 mins now and im no closer to finding the answer.

Question quoted from AQA M1 text book:

Telegraph poles, 40 m apart stand alongside a straight railway line. The times taken for a locomotoive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, repectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

Solve this equation ( I think) The idea is OK but I messed it up a bit. More in a mo

Using s=ut+1/2at²

At the second point s=40 t=2.5

40=2.5u+ 1/2(6.25a)

80=5u+ 6.25a ----(1)

At the 3rd Point s=80 t=2.3+2.5 t=4.8

80=4.8u+1/2(23.04a)

160=9.6u +23.04a ------(2)

Using the two equations to find u and a:

(1) x9.6 => 768=48u+60a

(2)x5 => 800=48u+115.2a

Equation (1)- Equation (2) => -32=-55.2a

a= -32/-55.2

a= 40/69 m/s

Substitute value for a back in to find u:

5u+6.25a=80

5u=80-6.25a

When a=40/69

5u= 80 - 3 43/69

u= (76 26/69)/5

u= 15 19/69

Whole thing seems abit wierd to me, so I probably messed up somewhere but I am sure thats the method.

At the second point s=40 t=2.5

40=2.5u+ 1/2(6.25a)

80=5u+ 6.25a ----(1)

At the 3rd Point s=80 t=2.3+2.5 t=4.8

80=4.8u+1/2(23.04a)

160=9.6u +23.04a ------(2)

Using the two equations to find u and a:

(1) x9.6 => 768=48u+60a

(2)x5 => 800=48u+115.2a

Equation (1)- Equation (2) => -32=-55.2a

a= -32/-55.2

a= 40/69 m/s

Substitute value for a back in to find u:

5u+6.25a=80

5u=80-6.25a

When a=40/69

5u= 80 - 3 43/69

u= (76 26/69)/5

u= 15 19/69

Whole thing seems abit wierd to me, so I probably messed up somewhere but I am sure thats the method.

Anujkhamar

i was wondering if you could help me??

i've bin stuck on a question for atleast 40 mins now and im no closer to finding the answer.

Question quoted from AQA M1 text book:

Telegraph poles, 40 m apart stand alongside a straight railway line. The times taken for a locomotoive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, repectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

i've bin stuck on a question for atleast 40 mins now and im no closer to finding the answer.

Question quoted from AQA M1 text book:

Telegraph poles, 40 m apart stand alongside a straight railway line. The times taken for a locomotoive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, repectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

Here is my second attempt. I am now checking to see if it's OK.

I have found an error. more in a mo Can you see my error ?

omg thank you!!!!!

yeh i see ure mistake in the first one steve, you wrote 2.3 where it was meant to be 2.5

once thank you all sooo much, you don't know how much peace you've put my mind at!

yeh i see ure mistake in the first one steve, you wrote 2.3 where it was meant to be 2.5

once thank you all sooo much, you don't know how much peace you've put my mind at!

Anujkhamar

omg thank you!!!!!

yeh i see ure mistake in the first one steve, you wrote 2.3 where it was meant to be 2.5

once thank you all sooo much, you don't know how much peace you've put my mind at!

yeh i see ure mistake in the first one steve, you wrote 2.3 where it was meant to be 2.5

once thank you all sooo much, you don't know how much peace you've put my mind at!

I made several errors. I think this gives the acceleration. Do you have the answer ? Put me out of misery !

mala2k

Using s=ut+1/2at²

At the second point s=40 t=2.5

40=2.5u+ 1/2(6.25a)

80=5u+ 6.25a ----(1)

At the 3rd Point s=80 t=2.3+2.5 t=4.8

80=4.8u+1/2(23.04a)

160=9.6u +23.04a ------(2)

Using the two equations to find u and a:

(1) x9.6 => 768=48u+60a

(2)x5 => 800=48u+115.2a

Equation (1)- Equation (2) => -32=-55.2a

a= -32/-55.2

a= 40/69 m/s

Substitute value for a back in to find u:

5u+6.25a=80

5u=80-6.25a

When a=40/69

5u= 80 - 3 43/69

u= (76 26/69)/5

u= 15 19/69

Whole thing seems abit wierd to me, so I probably messed up somewhere but I am sure thats the method.

At the second point s=40 t=2.5

40=2.5u+ 1/2(6.25a)

80=5u+ 6.25a ----(1)

At the 3rd Point s=80 t=2.3+2.5 t=4.8

80=4.8u+1/2(23.04a)

160=9.6u +23.04a ------(2)

Using the two equations to find u and a:

(1) x9.6 => 768=48u+60a

(2)x5 => 800=48u+115.2a

Equation (1)- Equation (2) => -32=-55.2a

a= -32/-55.2

a= 40/69 m/s

Substitute value for a back in to find u:

5u+6.25a=80

5u=80-6.25a

When a=40/69

5u= 80 - 3 43/69

u= (76 26/69)/5

u= 15 19/69

Whole thing seems abit wierd to me, so I probably messed up somewhere but I am sure thats the method.

I get the same acceleration as you... I haven't done the last bit.

- Numbers of Arrangements
- A level chemistry help please
- A Chain Sliding Through a Tube
- What units do I need for Edexcel IAL or IAS Further Maths
- the issac physics astronomy question
- Edexcel IAL M1 May 2024
- Device for 6th form and GCSEs
- Media Studies and Communication Student Laptop!!
- M1 pure maths
- Do colleges care about what modules I took for my IAL maths and further maths?
- Edexcel IAL Maths & Further Mathematics
- Weighing at Sea
- what ipad shall i get!
- Can i give IAL Edexcel Mathmematics across 3 sessions?
- Amount of Substance question
- Solomon or Elmwood or Delphis Papers?
- Health and social care Level 3 is stressing me out.
- Edexcel maths papers without blank pages in-between?
- How is IAL decision maths 1? Should I take it for CS?
- who have the answer of this paper?（Edexcel M1 2018 June）

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