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Equations of Motion - Constant Acceleration (M1 Mechanics Help)

Hello,
I am new to the forum. I have just started my A-Levels and am studying Biology, Chemistry, and Maths. Maths used to prove no difficulty for me in the past, but I am doing my homework for M1 now - Kinematics/Equation of Motion/Constant Acceleration, and I just need to be sent in the right direction of how to answer exam-style questions. I never liked Physics stuff but I like Maths, and I really like my Pure Maths.

So, one of the questions is:- M1 January 2009
"A train travels along a straight horizontal track with constant acceleration. Points A, B and C are on the track with B between A and C. The distance AB is 1200m and the distance BC is 2500m. As the train passes B, its speed is 26ms-1. The train takes 60s to travel from A to B."

(a) Find the speed of the train as it passes A. [3]
(b) Determine the acceleration of the train. [3]
(c) Calculate the speed of the train as it passes C, giving your answers to one decimal place. [3]

I would appreciate if you could show me how to answer, as well as providing the answer. I don't see my maths teacher until next week when the homework is due, so I would like help beforehand so I don't fall behind. I know to use the SUVAT equations but I don't know how to for this question.

I don't want to cheat my homework - only just to see how to answer. That's all. Thank you!

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Original post by ryanjj14
Hello,
I am new to the forum. I have just started my A-Levels and am studying Biology, Chemistry, and Maths. Maths used to prove no difficulty for me in the past, but I am doing my homework for M1 now - Kinematics/Equation of Motion/Constant Acceleration, and I just need to be sent in the right direction of how to answer exam-style questions. I never liked Physics stuff but I like Maths, and I really like my Pure Maths.

So, one of the questions is:- M1 January 2009
"A train travels along a straight horizontal track with constant acceleration. Points A, B and C are on the track with B between A and C. The distance AB is 1200m and the distance BC is 2500m. As the train passes B, its speed is 26ms-1. The train takes 60s to travel from A to B."

(a) Find the speed of the train as it passes A. [3]
(b) Determine the acceleration of the train. [3]
(c) Calculate the speed of the train as it passes C, giving your answers to one decimal place. [3]

I would appreciate if you could show me how to answer, as well as providing the answer. I don't see my maths teacher until next week when the homework is due, so I would like help beforehand so I don't fall behind. I know to use the SUVAT equations but I don't know how to for this question.

I don't want to cheat my homework - only just to see how to answer. That's all. Thank you!

Start by considering the motion from A to B only. Which of the SUVAT variables do you have for this motion?
Reply 2
Original post by Sir Cumference
Start by considering the motion from A to B only. Which of the SUVAT variables do you have for this motion?

S = 1200m,
T = 60s?

I'm not too sure here. That's why I asked for help.
Original post by ryanjj14
S = 1200m,
T = 60s?

I'm not too sure here. That's why I asked for help.

That's correct but you also have 'V' given to you in the question. Can you see what it is?

It's fine if you're not sure. I'm guiding you through the problem.
Reply 4
Original post by Sir Cumference
That's correct but you also have 'V' given to you in the question. Can you see what it is?

It's fine if you're not sure. I'm guiding you through the problem.

Would it be 26ms-1?

Thanks for helping me by the way. Much appreciated!
Original post by ryanjj14
Would it be 26ms-1?

Thanks for helping me by the way. Much appreciated!

Yes that's right. Now as soon as you have three of the SUVAT variables, that's enough to find all the others. Here you have

S = 1200
U = ?
V = 26
A = ?
T = 60

Ok so for part a) you need to find the SUVAT variable 'U' so the next step is to consider all of the SUVAT equations and find one that doesn't have 'A' in it (because we don't know that yet). Then you should be able to use the equation you choose to find 'U'. Does that help?
Reply 6
Original post by Sir Cumference
Yes that's right. Now as soon as you have three of the SUVAT variables, that's enough to find all the others. Here you have

S = 1200
U = ?
V = 26
A = ?
T = 60

Ok so for part a) you need to find the SUVAT variable 'U' so the next step is to consider all of the SUVAT equations and find one that doesn't have 'A' in it (because we don't know that yet). Then you should be able to use the equation you choose to find 'U'. Does that help?

Yes, it does help. Thanks a lot. I'll let you know how I get on!

Equation would be:- S=1/2(u+v)t, right?
Original post by ryanjj14
Yes, it does help. Thanks a lot. I'll let you know how I get on!

Equation would be:- S=1/2(u+v)t, right?

Yes that's the one you should use.
You could make an equation for a, which is (v-u)/t. You have v = 26 m s-1, t = 60 s, so a=(26 - u)/60
Now substitute that in s = ut + 0.5at^2
1200 = 60u + 0.5((26 - u)/60) × 60^2
1200 = 60u + ((26 - u)/120)×3600
1200 = 60u + (26 - u) × 30
1200 = 60u + 780 - 30u
u = (1200-780)/(60-30)
u = 14 m s-1
Then for finding a, just substitute the value of u.
And the final question, I think you could answer that now. (Answer is 40.9 m s-1)

P.S. I would suggest practising a lot cuz if you have trouble solving such a simple question.... it only gets harder thats all I'm saying. Sorry.
Reply 9
Original post by Sir Cumference
Yes that's the one you should use.

I've got an answer for U as 14ms-1. Would that be correct?

My steps are:
1200 = 1/2 (u+26) x 60
1200 = 30 (u+26)
1200 = 30u + 780
1200-780 = 30u
420 = 30u
420/30 = u
u = 14ms-1
Original post by SeverusESnape
X

There was no need to post the full answer. Let them do the work themselves with hints.
Original post by ryanjj14
I've got an answer for U as 14ms-1. Would that be correct?

My steps are:
1200 = 1/2 (u+26) x 60
1200 = 30 (u+26)
1200 = 30u + 780
1200-780 = 30u
420 = 30u
420/30 = u
u = 14ms-1

That's correct! Now try the next part.
Reply 12
Original post by Sir Cumference
That's correct! Now try the next part.

Okay, thanks so much.

For the acceleration, I had 0.2ms-2.

My method was:-
a = v-u / t
a = 26-14 / 60
a = 12 / 60
a = 0.2ms-2
Original post by Sir Cumference
There was no need to post the full answer. Let them do the work themselves with hints.

I'm sorry I didn't see all your posts. I'm new to this site literally created an account yesterday cuz had an m2 question. Could you help me with that btw?
Original post by ryanjj14
Okay, thanks so much.

For the acceleration, I had 0.2ms-2.

My method was:-
a = v-u / t
a = 26-14 / 60
a = 12 / 60
a = 0.2ms-2

That's also correct!
Reply 15
Original post by Sir Cumference
That's also correct!

Okay, thank you. For part c, I had 40.9ms-1 because it asked you to give your answer to 1 decimal place.

I used the formula: v^2 = u^2 + 2as
For SUVAT, I had s=2500m, u=26ms-1, v=?, a=0.2ms-2, t=?

I substituted them into the formula to find v and had the square root of 1676, which was 40.9ms-1.
Original post by ryanjj14
Okay, thank you. For part c, I had 40.9ms-1 because it asked you to give your answer to 1 decimal place.

I used the formula: v^2 = u^2 + 2as
For SUVAT, I had s=2500m, u=26ms-1, v=?, a=0.2ms-2, t=?

I substituted them into the formula to find v and had the square root of 1676, which was 40.9ms-1.

Correct.
Original post by SeverusESnape
I'm sorry I didn't see all your posts. I'm new to this site literally created an account yesterday cuz had an m2 question. Could you help me with that btw?


??
Reply 18
Original post by Sir Cumference
Correct.

Okay, thanks for helping me. I'll try another question now. If I don't get it, I'll post on here.

Thanks again! The next Q is a velocity-time graph.

Kind regards :-)
Original post by SeverusESnape
I'm sorry I didn't see all your posts. I'm new to this site literally created an account yesterday cuz had an m2 question. Could you help me with that btw?


Make a new thread with it if you want help on it

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