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    (Original post by B_9710)
    That's what I thought. The train could still keep decelerating to rest after the 6 seconds right?
    Yes it looks like it, especially since the first two parts sets you up to find the total time once you eliminate all the variables


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    (Original post by thefatone)
    it does but i see no other way than to use time since distance isn't given or mentioned at all in the question.
    ^^ i take it back
    maybe it's worth like a mark or something? or i've don't something horrendously wrong in which case
    Zacken TeeEm Student403
    i call for your help
    sorry I am teaching all day
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    (Original post by TeeEm)
    sorry I am teaching all day
    thought so >.>
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    (Original post by thefatone)
    it does but i see no other way than to use time since distance isn't given or mentioned at all in the question.
    ^^ i take it back
    maybe it's worth like a mark or something? or i've don't something horrendously wrong in which case
    Zacken TeeEm Student403
    i call for your help
    Yep, you've done this one wrong - sorry. It's fair simple, actually:

    Firs bit of information: s = 30, u = u, v = v, a = a, t = 2

    So 30 = 2u + 2a, so 15 = u+a

    Second bit of information: s = 30, initial velocity = u + 2a, v = , a = a, t=4

    So 30 = 4(u+2a) + 8a

    Then solve simul.
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    (Original post by Zacken)
    Yep, you've done this one wrong - sorry. It's fair simple, actually:

    Firs bit of information: s = 30, u = u, v = v, a = a, t = 2

    So 30 = 2u + 2a, so 15 = u+a

    Second bit of information: s = 30, initial velocity = u + 2a, v = , a = a, t=4

    So 30 = 4(u+2a) + 8a

    Then solve simul.
    all these c2 papers making me compile all my knowledge is kill me X_X
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    (Original post by Quido)
    Hmm, I just tried it again using simultaneous equations.
    I used s = ut + 1/2 at^2
    I got 30 = 2u - 2a and 60 = 6u - 18a and then solved them to get a = -2.5 and u = 17.5 m/s. And then using v = u + at I found t = 7.
    30 \neq 2(17.5) - 2(-2.5)
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    (Original post by Quido)
    Hmm, I just tried it again using simultaneous equations.
    I used s = ut + 1/2 at^2
    I got 30 = 2u - 2a and 60 = 6u - 18a and then solved them to get a = -2.5 and u = 17.5 m/s. And then using v = u + at I found t = 7.
    before setting up second stage, remember that the final velocity at stage 1 is the same as the initial velocity at stage 2




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    I got V as 12.5ms-1 and U as 17.5ms-1 (V is the speed after 2 seocnds has passed from t=0). But yes T is 7s to come to rest.
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    (Original post by B_9710)
    I got U as 12.5ms-1 and V as 17.5ms-1 (V is the speed after 2 seocnds has passed from t=0). But yes T is 7s to come to rest.
    I think you have used the right method but mixed up your U and V. Post workings? (sceenshot)
    The train is decelerating so how can it be possible for the final velocity to be higher than the initial?
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    (Original post by Zacken)
    30 \neq 2(17.5) - 2(-2.5)
    Sorry, I meant 30 = 2u + 2a
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    (Original post by Quido)
    The train is decelerating so how can it be possible for the final velocity to be higher than the initial?
    Other way round.
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    (Original post by B_9710)
    Other way round.
    But we know at the end V = 0 as the train is brought to rest.

    Also can I have help on this question.

    A ball is thrown vertically upwards at 25 m/s. Find the length of time for the ball is above 3m from the point of projection.

    s = 3
    u = 25
    v = ?
    a = 9.8
    t = ?

    I got v = 26.15 m/s using v^2 = u^2 + 2as and then got t = 0.117s which is time the ball is under 3m.
    How would I go about working out the time the ball is above 3m from there?
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    (Original post by Quido)
    But we know at the end V = 0 as the train is brought to rest.

    Also can I have help on this question.

    A ball is thrown vertically upwards at 25 m/s. Find the length of time for the ball is above 3m from the point of projection.

    s = 3
    u = 25
    v = ?
    a = 9.8
    t = ?

    I got v = 26.15 m/s using v^2 = u^2 + 2as and then got t = 0.117s which is time the ball is under 3m.
    How would I go about working out the time the ball is above 3m from there?
    Use s=ut+1/2at2 and you should form and solve the quadratic and get 2 values for t.Find the difference between the two t values and that is the required time T. Acceleration is acting in opposite direction to initial motion of the ball.

    (For the previous question I did not make V - the final speed/velocity - where the train had stopped. Please read my post again).
 
 
 
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