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Size:  296.7 KB hi for this question, I don't really understand the methods on the mark scheme. The method is to work out what t is and using simultaneous equations to solve for x. However, why do P and Q take the same amount of time to travel from O to B? The particles have different acceleration. So surely t would be different for the two particles. .. thanks.
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    (Original post by coconut64)
    Name:  1459608839809-1339212355.jpg
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Size:  296.7 KB hi for this question, I don't really understand the methods on the mark scheme. The method is to work out what t is and using simultaneous equations to solve for x. However, why do P and Q take the same amount of time to travel from O to B? The particles have different acceleration. So surely t would be different for the two particles. .. thanks.
    Yes they have different accelerations, but given that the questions says that they collide at B, they would have to be at the same place at the same time. Their timeline would be the same if that's what confused you, as the questions says both were released at the same time, just in different places

    So, now how could you generate two equations, given that know the point where they meet, the time when they meet (t) and both of their accelerations and initial velocities?
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    (Original post by KaylaB)
    Yes they have different accelerations, but given that the questions says that they collide at B, they would have to be at the same place at the same time. Their timeline would be the same if that's what confused you, as the questions says both were released at the same time, just in different places

    So, now how could you generate two equations, given that know the point where they meet, the time when they meet (t) and both of their accelerations and initial velocities?
    thanks for the reply. But p is released from A so surely Q has to travel for t s to O then T s to B. So the time taken would be greater for Q as it has to travel to o whereas p doesn't
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    (Original post by coconut64)
    thanks for the reply. But p is released from A so surely Q has to travel for t s to O then T s to B. So the time taken would be greater for Q as it has to travel to o whereas p doesn't
    It's not like P moves to B and then sits around like a sentient human beinf waiting for Q to come along an hit it. P starts from O and moves (from rest) at an acceleration of 3 ms^-1.

    Q starts from A but has a higher initial velocity and acceleration 4 ms^-1 so Q and P are moving in the same direction and point B is when Q finally catches up to P and collides with it. This happens at time t. Hence, at time t, Q has travelled 20 + x metres where as P has travelled just x metres.

    You cannot say they took different times. Q collides with P at a time t means that P collides with Q at a time t as well... it cannot be that Q collides with P at time t and then P collides with Q at another time... that disobeys the definition of the word "collide".
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    Apologies for any errors or mis-writings, I'm on my phone and not home so typing is dodgy.
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    (Original post by coconut64)
    thanks for the reply. But p is released from A so surely Q has to travel for t s to O then T s to B. So the time taken would be greater for Q as it has to travel to o whereas p doesn't
    Q has an acceleration of 4ms^-2, whereas P has an acceleration of 3ms^-2
    So Q is travelling faster behind P, the slower object

    For example if you see someone leave their house 10m in front of you and start walking behind them (albeit creepy), if you were walking them and you walked quite quickly, and the person in front of you travelled rather slowly you would catch up with them or bump into each other at the same time even though you both set off at different places.
    (If that makes sense, or I've just confused you even more, I've gone to deep to abandon this)
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    (Original post by KaylaB)
    Q has an acceleration of 4ms^-2, whereas P has an acceleration of 3ms^-2
    So Q is travelling faster behind P, the slower object

    For example if you see someone leave their house 10m in front of you and start walking behind them (albeit creepy), if you were walking them and you walked quite quickly, and the person in front of you travelled rather slowly you would catch up with them or bump into each other at the same time even though you both set off at different places.
    (If that makes sense, or I've just confused you even more, I've gone to deep to abandon this)
    Thanks for the analogy. I think I get it now. U didn't confuse me dw.
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    (Original post by Zacken)
    It's not like P moves to B and then sits around like a sentient human beinf waiting for Q to come along an hit it. P starts from O and moves (from rest) at an acceleration of 3 ms^-1.

    Q starts from A but has a higher initial velocity and acceleration 4 ms^-1 so Q and P are moving in the same direction and point B is when Q finally catches up to P and collides with it. This happens at time t. Hence, at time t, Q has travelled 20 + x metres where as P has travelled just x metres.

    You cannot say they took different times. Q collides with P at a time t means that P collides with Q at a time t as well... it cannot be that Q collides with P at time t and then P collides with Q at another time... that disobeys the definition of the word "collide".
    I get it now. Thanks
 
 
 
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