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    • Thread Starter
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    Hi,

    Could someone please try this question and tell me what they get?

    For part i) I get t=100

    and part ii) I get 400m

    Thanks

    Hawk
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    I get the same.
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    Taking A:
    a(t)=0.4
    v(t)=0.4t+c and as it starts from rest c=0 so v(t)=0.4t
    x(t)=0.2t^2+d and as we are taking this place as the origin, d=0 so, for a:
    X(T)=0.2t^2

    For B:
    a(t)=0.2
    v(t)=0.2t+6
    x(t)0.1t^2+6t+400

    When they meet:
    0.2t^2=0.1t^2+6t+400
    0.1t^2-6t-400=0
    t^2-60t-4000=0
    t=(60+(3600+16000)^0.5)/2
    =100s (as when you take away the square root, you get a negative value of t and t>0 for this)

    When t=100, A is 0.2x100^2 from where it started (2km) and it started 2.4km from P so it is 0.4km from P when the lorries draw level.

    So yes, I agree.
    • Thread Starter
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    ty,

    it was a question on todays M1 paper today btw (CCEA)
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    I was gonna say - thought all the M1 papers were over by now
 
 
 
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