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    (Original post by wat a wizard)
    Can anyone help me with 4)ii) ? Isn't the area under the graph the distance travelled?

    paper ; https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf

    markscheme;

    https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf
    Yep. Area under v,t graph is the distance travelled.
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    (Original post by MsFahima)
    Yep. Area under v,t graph is the distance travelled.
    edited
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    (Original post by wat a wizard)
    edited
    So you don't need any help right?
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    (Original post by MsFahima)
    So you don't need any help right?
    yeah, thanks
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    Anyone really struggling with this?I would advise u to check out examsolutions,Whole year at school i barely could do a question due to teacher not good and basically he did full m1 in 3-4 months and then we forcused on c3.I looked through examsolutions and basically what he goes through is exqctly wat u need.Now im close to getting A's.Well thats my story to shed light to yall lol.I did m1 just 2 weeks ago the full lot bymyself.Anyways GL all and ull be find!!
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    hi can anyone help me with m1 jan 06 q3iii?

    A motorcyclist starts from rest at a point O and travels in a straight line. His velocity after t secondsis vm s−1, for 0 ≤ t ≤ T, where v = 7.2t − 0.45t2. The motorcyclist’s acceleration is zero when t = T.

    (i) Find the value of T. [4]
    (ii) Show that v = 28.8 when t = T. [1]

    For t ≥ T the motorcyclist travels in the same direction as before, but with constant speed 28.8 m s−1.

    (iii) Find the displacement of the motorcyclist from O when t = 31.

    (the answer is 816m)
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    (Original post by steven123456789)
    hi can anyone help me with m1 jan 06 q3iii?

    A motorcyclist starts from rest at a point O and travels in a straight line. His velocity after t secondsis vm s−1, for 0 ≤ t ≤ T, where v = 7.2t − 0.45t2. The motorcyclist’s acceleration is zero when t = T.

    (i) Find the value of T. [4]
    (ii) Show that v = 28.8 when t = T. [1]

    For t ≥ T the motorcyclist travels in the same direction as before, but with constant speed 28.8 m s−1.

    (iii) Find the displacement of the motorcyclist from O when t = 31.

    (the answer is 816m)
    (i) you need to get an equation for acceleration by differentiating the velocity equation. a = dv/dt = 7.2 - 0.9t. Put this equal to 0 to get out T=8.
    (ii) Put T=8 into the velocity equation to get out v=28.8.
    (iii) For displacement, you want to integrate the velocity equation, to get s = 3.6t2 - 0.15t3 (+c but c=0 as when t=0, s=0). Then put t=8 into this to find the distance travelled in the first 8 seconds (153.6). After this, you can use s=ut as it's constant velocity where t=31-8 = 23. Therefore s travelled at 8<T<31 = 28.8*23 = 662.4. Total distance is 153.6+662.4=816m.

    Hope that made some sort of sense, I'm not the best at explaining :lol:
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    (Original post by kawehi)
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    Right so I thought I would attempt a paper and then I gave up.

    Do you mind helping me with a few questions...

    Question 1 first of all

    Care to explain it. Thanks
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    (Original post by Super199)
    Right so I thought I would attempt a paper and then I gave up.

    Do you mind helping me with a few questions...

    Question 1 first of all

    Care to explain it. Thanks
    Which paper is this? And of course, no problem!
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    (Original post by kawehi)
    Which paper is this? And of course, no problem!
    oops sorry about that. It is Jan 2007.
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    (Original post by chloe-jessica)
    (i) you need to get an equation for acceleration by differentiating the velocity equation. a = dv/dt = 7.2 - 0.9t. Put this equal to 0 to get out T=8.
    (ii) Put T=8 into the velocity equation to get out v=28.8.
    (iii) For displacement, you want to integrate the velocity equation, to get s = 3.6t2 - 0.15t3 (+c but c=0 as when t=0, s=0). Then put t=8 into this to find the distance travelled in the first 8 seconds (153.6). After this, you can use s=ut as it's constant velocity where t=31-8 = 23. Therefore s travelled at 8<T<31 = 28.8*23 = 662.4. Total distance is 153.6+662.4=816m.

    Hope that made some sort of sense, I'm not the best at explaining :lol:
    thank you!!!
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    (Original post by Super199)
    oops sorry about that. It is Jan 2007.
    1)i)

    We are given the tension in the tow bar, 700N. This is what is pulling on the trailer. If we think about the DOM of the trailer, we can find the net force.

    700 - drag force (D)

    Using F=ma

    700 - D = 600 x 0.8
    D = 220

    ii) See if you can do this one, using the same exact principles as the last part! draw a diagram and see what forces you can fill in.
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    (Original post by kawehi)
    1)i)

    We are given the tension in the tow bar, 700N. This is what is pulling on the trailer. If we think about the DOM of the trailer, we can find the net force.

    700 - drag force (D)

    Using F=ma

    700 - D = 600 x 0.8
    D = 220

    ii) See if you can do this one, using the same exact principles as the last part! draw a diagram and see what forces you can fill in.
    I am a bit confused sorry. Do you mind attaching a diagram
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    (Original post by Super199)
    I am a bit confused sorry. Do you mind attaching a diagram
    Name:  3m1.png
Views: 191
Size:  1.2 KB
    If you only consider the trailer, look at the different forces that are acting on it, then try to work out the resultant force. (for part i)
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    (Original post by kawehi)
    Name:  3m1.png
Views: 191
Size:  1.2 KB
    If you only consider the trailer, look at the different forces that are acting on it, then try to work out the resultant force. (for part i)
    ah okay I get part i.

    Not sure with ii,
    So now we 2100N from the car.
    F=ma
    2100-700-resistance = 1100*0.8 (Is what is in the ms)
    Why is it this.
    Why isn't it just 2100-(700+220)
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    (Original post by Super199)
    ah okay I get part i.

    Not sure with ii,
    So now we 2100N from the car.
    F=ma
    2100-700-resistance = 1100*0.8 (Is what is in the ms)
    Why is it this.
    Why isn't it just 2100-(700+220)
    The drag force was just for the trailer. When you do these sort of 'two attached particle' problems, you have to consider them completely individually. So the car has a whole new drag force, completely independent from the trailer.
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    (Original post by kawehi)
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    Check the opening post please.. is there any other important tips I should add? Thanks!
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    (Original post by kawehi)
    The drag force was just for the trailer. When you do these sort of 'two attached particle' problems, you have to consider them completely individually. So the car has a whole new drag force, completely independent from the trailer.
    ah I see. So I shouldn't assume it will be the same as the previous part.

    The next question is 4iia.
    For the part above that I worked out the speed of L after the collision to be 1ms^-1.

    It then says M collides with N. From looking at the mark scheme it changes back to using 2ms^-1. Why is that?
    If you could explain that part it would be appreciated.

    Thanks
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    (Original post by MsFahima)
    Check the opening post please.. is there any other important tips I should add? Thanks!
    Haha the linked to answered questions, 90% of them are probably the ones I couldn't do. FML :P
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    (Original post by Super199)
    Haha the linked to answered questions, 90% of them are probably the ones I couldn't do. FML :P
    Haha. It's okay. Better to ask now than struggle in the exam! :P
 
 
 

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