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    Hi guys, Am I right in saying... Whether velocity has a negative sign or a positive sign is dependent on the direction of motion. For example, if you throw a ball up into the air from 3m above the ground (taking positive direction as upwards) then your initial velocity is 3m per second. However,if you throw a ball up into the air from 3m above the ground (taking positive direction as downwards) then your initial velocity is -3m per second.

    Also, for this question part iii only, I decided to tackle it using 2 different ways. Could someone please help me confirm that I am doing this the right way? # It's related to the question I am asking above.
    My working out is in the doc...So in my working out, am I right to put u=10ms^-1 in way one and put u=-10ms^-1 in way two?
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    Anyone please? It's just about my working out for part iii) and the general question I have in the beginning..
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    (Original post by sienna2266)
    Hi guys, Am I right in saying... Whether velocity has a negative sign or a positive sign is dependent on the direction of motion. For example, if you throw a ball up into the air from 3m above the ground (taking positive direction as upwards) then your initial velocity is 3m per second. However,if you throw a ball up into the air from 3m above the ground (taking positive direction as downwards) then your initial velocity is -3m per second.
    Yes.

    (Original post by sienna2266)
    Also, for this question part iii only, I decided to tackle it using 2 different ways. Could someone please help me confirm that I am doing this the right way? # It's related to the question I am asking above.
    Both ways are fine. s represents displacement, which is a vector quantity, so it can very well be -ve in any case.
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    Okay so if you’re familiar with suvat, find the ones you have and are trying to find.

    The resistive forces change (increase for B) which means the acceleration changes. Find the new acceleration during the fault period before the train comes to an end

    S = ?, u = 10 (initial from start period), v = 0 (comes to an end), a = new found
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    (Original post by RDKGames)
    Yes.
    Both ways are fine. s represents displacement, which is a vector quantity, so it can very well be -ve in any case.
    Thanks so much for your reply. Just one thing about (iiii):
    We are given in the question that the fault occurs when the train is travelling at 10ms^-1 so therefore the value for the initial velocity u is 10. In my working out below, I have chosen to work with the acceleration on the left, so I have chosen the direction of motion of the train to be to the left. Also, since I have chosen the positive direction as the right, this means u=-10. If I had chosen the positive direction as the left, then u would have been 10ms^-1. Is this correct?

    Should u = -10ms^-1 and not 10ms^-1 below?
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    (Original post by RDKGames)
    Yes.
    Both ways are fine. s represents displacement, which is a vector quantity, so it can very well be -ve in any case.
    Sorry to be so annoying.
    Just one thing about (iiii):
    We are given in the question that the fault occurs when the train is travelling at 10ms^-1 so therefore the value for the initial velocity u is 10. In my working out below, I have chosen to work with the acceleration on the left, so I have chosen the direction of motion of the train to be to the left. Also, since I have chosen the positive direction as the right, this means u=-10. If I had chosen the positive direction as the left, then u would have been 10ms^-1. Is this correct?

    Should u = -10ms^-1 and not 10ms^-1 below?
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    Thanks so much for the help so far btw!
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    (Original post by sienna2266)
    ...
    Didn't pick up on this as I skimmed through it at first, but since the train is going to the right, if you're taking LEFT as the +ve direction, then u=-10 and a=0.05. If you're taking RIGHT as the +ve direction, u=10 and a=-0.05.
    All you need to do in your sheet to have it correct is swap the signs of u in both ways of working out. This doesn't affect the answer though, obviously, since u is squared.
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    (Original post by RDKGames)
    Didn't pick up on this as I skimmed through it at first, but since the train is going to the right, if you're taking LEFT as the +ve direction, then u=-10 and a=0.05. If you're taking RIGHT as the +ve direction, u=10 and a=-0.05.
    All you need to do in your sheet to have it correct is swap the signs of u in both ways of working out. This doesn't affect the answer though, obviously, since u is squared.
    Ohh!!
    Oh so the velocity is dependent on the direction of motion ( so the way the train is going)?
    And the velocity is not dependent on the direction of acceleration I have chosen?
    So given that the train had a driving force to the right in previous parts of the question, we know the train is heading to the right or I mean slowing down to the right. Let's say I take right as the positive direction and left as the direction of acceleration. This means the direction of motion is in the positive direction as it's to the right so anything going against it will be negative. The acceleration is going against the direction of motion so it is negative. However, the velocity always goes with the direction of motion in any scenario,so the initial velocity is positive.
    Thanks so much - would really appreciate it if you could please let me know if I got it right this time
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    (Original post by sienna2266)
    ...
    Yes, you got it.
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    (Original post by RDKGames)
    Yes, you got it.
    Yay! Thanks so much
 
 
 
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