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    I'm having trouble with this problem. I've tried to solve it, but I feel like I'm going around in circles. Could anyone help?

    'Theresa is handing the baton on to Magda in a relay race. Theresa is running at a constant speed and when she is 4.5 m away Magda starts running with acceleration 1 m s-2. Theresa continues at a constant speed and just manages to catch up and hand over the baton. How fast was Theresa running?'
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    (Original post by fablereader)
    I'm having trouble with this problem. I've tried to solve it, but I feel like I'm going around in circles. Could anyone help?

    'Theresa is handing the baton on to Magda in a relay race. Theresa is running at a constant speed and when she is 4.5 m away Magda starts running with acceleration 1 m s-2. Theresa continues at a constant speed and just manages to catch up and hand over the baton. How fast was Theresa running?'
    draw a speed vs time graph up to the point where Magda is running at  \displaystyle v \: \text{ms}^{-1}. ie  \displaystyle \text{t=}v \:\text{secs}. this is because after  \displaystyle \text{t=v} magda will be running at a speed more than v, therefore there's no chance theresa will catch her

    then find areas under graphs and say that distance theresa ran + 4.5 = distance magda ran and youll find v
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    I did it by using a couple of equations, but I'm not sure if it's right.
    After time T Theresa will have run 4.5 m. For this same length of time, Madga? whoever it was will have been accelerating to the same speed as Theresa, so that she may catch up.
    4.5/V = T
    at = V

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    (Original post by DylanJ42)
    draw a speed vs time graph up to the point where Magda is running at  \displaystyle v \: \text{ms}^{-1}. ie  \displaystyle \text{t=}v \:\text{secs}. this is because after  \displaystyle \text{t=v} magda will be running at a speed more than v, therefore there's no chance theresa will catch her

    then find areas under graphs and say that distance theresa ran + 4.5 = distance magda ran and youll find v
    I'm sorry, but I don't get it. I do know about finding the area under a line in a velocity-time graph, but I don't get how I can use it in this situation.
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    (Original post by fablereader)
    I'm sorry, but I don't get it. I do know about finding the area under a line in a velocity-time graph, but I don't get how I can use it in this situation.
    no no don't be sorry, my explanation was maybe too vague.

    here is a diagram;

    Name:  timespeed.jpg
Views: 81
Size:  28.0 KB

    time = 0 is the instant the scenario starts, we don't need to worry about anything before that.

    let theresa have constant speed 'v' ms-1, this means no matter what the time is theresa will be running at 'v'

    now, since magda is running with an acceleration of 1 ms-2 this means that at;

    t = 0, v = 0
    t = 1, v = 1
    t = 2, v = 2

    and most importantly, at t=v, v = v

    Now this point is important because once magda starts running at speeds more than v, it will be physically impossible for theresa to catch up since madga will be running faster.

    Since the question tells you that theresa just catches magda this means she reaches madga at time = v, ie the last possible second before magda starts running at a faster speed than theresa.

    got everything so far?


    so now we find the area under both curves/lines from time = 0 to time = v.

    However, on top of this, we must remember that theresa has to run an extra 4.5m in this time period than magda to catch her.

    So "distance theresa runs + 4.5 = distance magda runs" between t=0 and t=v

    re wording this slightly we get to;

     \DISPLAYSTYLE \text{Area under theresa's curve + 4.5 = Area under magda's curve , between t=0 and t=v}
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    (Original post by DylanJ42)
    no no don't be sorry, my explanation was maybe too vague.

    here is a diagram;

    Name:  timespeed.jpg
Views: 81
Size:  28.0 KB

    time = 0 is the instant the scenario starts, we don't need to worry about anything before that.

    let theresa have constant speed 'v' ms-1, this means no matter what the time is theresa will be running at 'v'

    now, since magda is running with an acceleration of 1 ms-2 this means that at;

    t = 0, v = 0
    t = 1, v = 1
    t = 2, v = 2

    and most importantly, at t=v, v = v

    Now this point is important because once magda starts running at speeds more than v, it will be physically impossible for theresa to catch up since madga will be running faster.

    Since the question tells you that theresa just catches magda this means she reaches madga at time = v, ie the last possible second before magda starts running at a faster speed than theresa.

    got everything so far?


    so now we find the area under both curves/lines from time = 0 to time = v.

    However, on top of this, we must remember that theresa has to run an extra 4.5m in this time period than magda to catch her.

    So "distance theresa runs + 4.5 = distance magda runs" between t=0 and t=v

    re wording this slightly we get to;

     \DISPLAYSTYLE \text{Area under theresa's curve + 4.5 = Area under magda's curve , between t=0 and t=v}
    I get all that, thanks. However, doing the math I get v2=-9, which then delves into complex numbers (which weren't even explored in the main Maths module, and shouldn't be used in Mechanics). Am I missing something?
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    (Original post by fablereader)
    I get all that, thanks. However, doing the math I get v2=-9, which then delves into complex numbers (which weren't even explored in the main Maths module, and shouldn't be used in Mechanics). Am I missing something?
    can you post workings, i got to  \displaystyle v^2 = +9
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    (Original post by DylanJ42)
    can you post workings, i got to  \displaystyle v^2 = +9
    Area under Theresa's line is v2, and the area under Magda's is v2/2.

    v2 + 4.5 = v2/2
    Multiply by 2
    2v2 + 9 = v2
    Subtract 9 and v2 from both sides
    v2 = -9.

    This is my workings (thanks for checking in so late).
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    (Original post by fablereader)
    Area under Theresa's line is v2, and the area under Magda's is v2/2.

    v2 + 4.5 = v2/2
    Multiply by 2
    2v2 + 9 = v2
    Subtract 9 and v2 from both sides
    v2 = -9.

    This is my workings (thanks for checking in so late).
    yea i released I made a mistake, let me redo it, sorry
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    (Original post by fablereader)
    Area under Theresa's line is v2, and the area under Magda's is v2/2.

    v2 + 4.5 = v2/2
    Multiply by 2
    2v2 + 9 = v2
    Subtract 9 and v2 from both sides
    v2 = -9.

    This is my workings (thanks for checking in so late).
    after a little think ive concluded that im retarded...


    since theresa has to run 4.5metres more then the equation is actually

    (Original post by DylanJ42)

     \DISPLAYSTYLE \text{Area under theresa's curve = Area under magda's curve + 4.5 , between t=0 and t=v}
    since theresa is the one running the longer distance.

    (in my workings I made two mistakes with signs which accidentally made me right, im super sorry about that)

    Edit: yea that actually took me 10+ minutes to realise, this is why im not doing any mechanics past M2 :laugh:
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    (Original post by DylanJ42)
    after a little think ive concluded that im retarded...


    since theresa has to run 4.5metres more then the equation is actually



    since theresa is the one running the longer distance.

    (in my workings I made two mistakes with signs which accidentally made me right, im super sorry about that)

    Edit: yea that actually took me 10+ minutes to realise, this is why im not doing any mechanics past M2 :laugh:
    You're not the only one who missed the obvious. Doing the math, the answer is 3 m s-1 (the answers at the back confirm). Thank you so much for your help!
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    (Original post by fablereader)
    You're not the only one who missed the obvious. Doing the math, the answer is 3 m s-1 (the answers at the back confirm). Thank you so much for your help!
    my brain is not made for mechanics :laugh: its no problem sorry again for the mistake
 
 
 
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