need help on an m1 question

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.
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.

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.
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.

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 ?
draw a velocity against time graph, assume constant acceleration. you know the area under the curve...
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!
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!

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.

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