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1. Hi guys I'm really struggling on this Mechanics question. I would really appreciate it if you could help me out.

Two trains A and B run on parallel straight tracks. Initially both are at rest in a station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 s up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measured in seconds. It accelerated with the same initial acceleration as A, up to a speed of 60m/s. It then moves at a constant speed of 60m/s. It thenmoes at a constant speed of 60m/s. Train B overtakes A after both trains have reached their maximum speed. Train B overtakes A when t=T.

a). Sketch, on the same diagram, the speed-time graphs of both trains for 0≤t≤T. (3 marks)

b). Find the value of T. (9 marks)
2. (Original post by alfinkjoseph)
Hi guys I'm really struggling on this Mechanics question. I would really appreciate it if you could help me out.

Two trains A and B run on parallel straight tracks. Initially both are at rest in a station and level with each other. At time t=0, A starts to move. It moves with constant acceleration for 12 s up to a speed of 30m/s, and then moves at a constant speed of 30m/s. Train B starts to move in the same direction as A when t=40, where t is measured in seconds. It accelerated with the same initial acceleration as A, up to a speed of 60m/s. It then moves at a constant speed of 60m/s. It thenmoes at a constant speed of 60m/s. Train B overtakes A after both trains have reached their maximum speed. Train B overtakes A when t=T.

a). Sketch, on the same diagram, the speed-time graphs of both trains for 0≤t≤T. (3 marks)

b). Find the value of T. (9 marks)
How far have you got?

The gradient of a speed-time graph gives the acceleration.
3. (Original post by ombtom)
How far have you got?

The gradient of a speed-time graph gives the acceleration.
I've found that the acceleration is 2.5 m/s2
4. (Original post by alfinkjoseph)
I've found that the acceleration is 2.5 m/s2
That's right.

a = (v-u)/t

Rearrange this for t, and use v=60 to find the time that B reaches 60m/s.
5. (Original post by ombtom)
That's right.

a = (v-u)/t

Rearrange this for t, and use v=60 to find the time that B reaches 60m/s.
Is the graph I drew for part 1 right or not
6. (Original post by ombtom)
That's right.

a = (v-u)/t

Rearrange this for t, and use v=60 to find the time that B reaches 60m/s.
What would u be in this situation.
Is that all that is needed because the question is worth 9 marks.
7. (Original post by alfinkjoseph)
Is the graph I drew for part 1 right or not

For part b):

Distance = speed * time = area under the graph.

When B overtakes A, they will have travelled the same distance, so the area under line A will equal the area under line B.
8. (Original post by alfinkjoseph)
What would u be in this situation.
Is that all that is needed because the question is worth 9 marks.
That was for part a). B started at rest, so u = 0.
9. (Original post by ombtom)
That was for part a). B started at rest, so u = 0.
So is T=24 seconds
10. (Original post by alfinkjoseph)
So is T=24 seconds
No, that is still for part a) (T is the time at which B overtakes A). Train B reaches maximum speed 24 seconds after it started to accelerate.
11. (Original post by ombtom)
No, that is still for part a). Train B reaches maximum speed 24 seconds after it started to accelerate.
So does train B start accelerating at 40 seconds and reaches top speed of 60m/s at 64 seconds
12. (Original post by alfinkjoseph)
So does train B start accelerating at 40 seconds and reaches top speed of 60m/s at 64 seconds
Yes! I think you need to mark that on your x-axis as it is required for part b).
13. (Original post by ombtom)
Yes! I think you need to mark that on your x-axis as it is required for part b).
What steps would you then need to do to find out the value of T.
14. (Original post by alfinkjoseph)
What steps would you then need to do to find out the value of T.
Distance = speed * time = area under the graph.

At the time when B overtakes A, they will have travelled the same distance, so the area under line A will equal the area under line B. I suggest that you use the formula for the area of a trapezium, rather than triangles and rectangles.
15. (Original post by ombtom)
Distance = speed * time = area under the graph.

At the time when B overtakes A, they will have travelled the same distance, so the area under line A will equal the area under line B. I suggest that you use the formula for the area of a trapezium, rather than triangles and rectangles.
I'm getting different areas for both A and B. Can you tell me how to get the area for both of them pls.
16. (Original post by alfinkjoseph)
I'm getting different areas for both A and B. Can you tell me how to get the area for both of them pls.
Find them in terms of T, e.g. the area under B = 1/2 * 60 * (T - 40 + T - 64), using 'area = half x distance between the parallel sides x sum of the parallel sides'.

... This simplifies to B = 30 * (2T - 104).

Repeat for the area under A, equate them, and then solve for T.
17. (Original post by ombtom)
Find them in terms of T, e.g. the area under B = 1/2 * 60 * (T - 40 + T - 64), using 'area = half x distance between the parallel sides x sum of the parallel sides'.

... This simplifies to B = 30 * (2T - 104).

Repeat for the area under A, equate them, and then solve for T.
Is area of A = 30T-180
18. (Original post by alfinkjoseph)
Is area of A = 30T-180
Area under line A = 1/2 * 30 * (T + T - 12)
= 15 * (2T - 12)
= 30T - 180.

... Yes.
19. (Original post by ombtom)
Area under line A = 1/2 * 30 * (T + T - 12)
= 15 * (2T - 12)
= 30T - 180.

... Yes.
So would T=98 seconds
20. (Original post by alfinkjoseph)
So would T=98 seconds
I also got T = 98 seconds, so hopefully that's right. It seems reasonable; it is after both trains have reached their maximum speeds (as stated in the question).

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