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    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)
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    (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.
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    (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
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    (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.
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    (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.
    Name:  2015-11-02 17.55.31.jpg
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Size:  495.5 KBIs the graph I drew for part 1 right or not
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    (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.
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    (Original post by alfinkjoseph)
    Name:  2015-11-02 17.55.31.jpg
Views: 508
Size:  495.5 KBIs the graph I drew for part 1 right or not
    Your graph is correct.

    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.
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    (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.
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    (Original post by ombtom)
    That was for part a). B started at rest, so u = 0.
    So is T=24 seconds
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    (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.
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    (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
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    (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).
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    (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.
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    (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.
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    (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.
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    (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.
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    (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
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    (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.
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    (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
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    (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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