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    Could you attempt the questions yourself? I just want to check my answers. If you can that would be very appreciated. Thanks.

    *PIC BELOW*

    My answers:

    (a) 60m/s
    (b) 450m
    (c) 3000N
    (d) 30000kgm/s
    (e) 450000J
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    thanks again
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    (Original post by KINGYusuf)
    thanks again
    I agree with all of them.
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    (Original post by RDKGames)
    I agree with all of them.
    What about this one?

    Sorry about annoying you, just want everything to be right so I can confirm I understand everything!

    I got:

    (a) 22m/s
    (b) 52m
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    (Original post by KINGYusuf)
    What about this one?

    Sorry about annoying you, just want everything to be right so I can confirm I understand everything!

    I got:

    (a) 22m/s
    (b) 52m
    I agree with the first one but not the second one. What was your calculation?
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    (Original post by RDKGames)
    I agree with the first one but not the second one. What was your calculation?
    Oops used the wrong velocity. I meant 36m. Is that correct?
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    (Original post by KINGYusuf)
    Oops used the wrong velocity. I meant 36m. Is that correct?
    Nope. What's your calculation??
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    (Original post by RDKGames)
    Nope. What's your calculation??
    s = ut + 1/2at^2

    I'm a bit confused. Is it saying to calculate the distance from the braking part only? Or the whole distance? Because then t would be different.
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    (Original post by KINGYusuf)
    s = ut + 1/2at^2

    I'm a bit confused. Is it saying to calculate the distance from the braking part only? Or the whole distance? Because then t would be different.
    You don't use that formula because that gives you the displacement and not the distance traveled. Displacement is how much it has moved from its starting position. If I walk a full square of side length 1m then my distance traveled is 4m while my displacement is 0m as I am back where I started.

    If you draw a time-velocity graph you can easily work out the distance traveled as it is the area under the graph. Also, it is refering to the full time of motion, so for the full 12 seconds.
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    (Original post by RDKGames)
    You don't use that formula because that gives you the displacement and not the distance traveled. Displacement is how much it has moved from its starting position. If I walk a full square of side length 1m then my distance traveled is 4m while my displacement is 0m as I am back where I started.

    If you draw a time-velocity graph you can easily work out the distance traveled as it is the area under the graph. Also, it is refering to the full time of motion, so for the full 12 seconds.
    ahhh.

    My attempt:

    I'm not the best at understanding and drawing graphs but does it look something like this?

    The answer I got was 202m. What a weird answer, I defo got this wrong :/

    Could you tell me the answer anyway?
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    (Original post by KINGYusuf)
    ahhh.

    My attempt:

    I'm not the best at understanding and drawing graphs but does it look something like this?

    The answer I got was 202m. What a weird answer, I defo got this wrong :/

    Could you tell me the answer anyway?
    I agree with that answer, your diagram is correct. Though you should label the axis if your graph is to be marked in questions like these.
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    (Original post by RDKGames)
    I agree with that answer, your diagram is correct. Though you should label the axis if your graph is to be marked in questions like these.
    Thank you for everything! Much appreciated
 
 
 
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