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    Hi guys, so I get part i) but unsure about ii).i) is 3pm, ii) is 150km
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    Here is my diagram for ii)
    Attachment 726526726528
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    (Original post by sienna2266)
    Hi guys, so I get part i) but unsure about ii).i) is 3pm, ii) is 150km
    In part (ii) the plane should be travelling east rather than west. The wind direction looks OK (from the SW, towards the NE). But you can simplify part (ii) by considering that displacement of the plane away from Prague is solely due to two hours-worth of wind towards the NE.
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    (Original post by old_engineer)
    In part (ii) the plane should be travelling east rather than west. The wind direction looks OK (from the SW, towards the NE). But you can simplify part (ii) by considering that displacement of the plane away from Prague is solely due to two hours-worth of wind towards the NE.
    Thank you very much for your reply!
    Just corrected my diagram, is it correct now?
    Attachment 726560
    I am not sure what would be the next step from here?
    Many thanks
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    for part iii) you want the resultant of the wind & still air vectors to be in the direction East.
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    215 knots is kinda slow.
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    (Original post by the bear)
    for part iii) you want the resultant of the wind & still air vectors to be in the direction East.
    Thanks for you reply! I am still confused with part ii) though. Here is my diagram:
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    What would be the next step?
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    (Original post by sienna2266)
    Thanks for you reply! I am still confused with part ii) though. Here is my diagram:
    What would be the next step?
    Work out the resultant velocity.

    The question basically asks in part (i) how long does the normal journey takes in the simplest case with no resistant forces. T=3 as found.

    In part (ii), you are considering the journey of length 3 hours, where for the first hour the plane flies as usual. We are now interested to see where it ends up after 2 hours going at the resultant velocity. So the next step is to determine how much distance the plane covers at speed R going for 2 hours. Also, you wish to determine the bearing of R to see in what direction the plane is offset from the original course.
    Once you have direction and speed of the resultant path, simply consider drawing a displacement diagram between two horizontal points H and P 1200km apart representing the airports, and now draw on the path that the plane takes using the info previously. Label the point it ends up at as A. You wish to determine distance AP. Consider cosine rule.
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    (Original post by sienna2266)
    Thanks for you reply! I am still confused with part ii) though. Here is my diagram:
    Name:  70.PNG
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    What would be the next step?
    The new velocity diagram is correct. Now consider that the plane is already 400km due east of Heathrow when this diagram comes into play, as the first hour was wind-free. Prague lies a further 800km due east. Where will the plane be after a further two hours of flying? It will be a further 800km east plus 150km north-east (your vectors times two hours). It’s displacement from Prague is then...
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    (Original post by RDKGames)
    Work out the resultant velocity.

    The question basically asks in part (i) how long does the normal journey takes in the simplest case with no resistant forces. T=3 as found.

    In part (ii), you are considering the journey of length 3 hours, where for the first hour the plane flies as usual. We are now interested to see where it ends up after 2 hours going at the resultant velocity. So the next step is to determine how much distance the plane covers at speed R going for 2 hours. Also, you wish to determine the bearing of R to see in what direction the plane is offset from the original course.
    Once you have direction and speed of the resultant path, simply consider drawing a displacement diagram between two horizontal points H and P 1200km apart representing the airports, and now draw on the path that the plane takes using the info previously. Label the point it ends up at as A. You wish to determine distance AP. Consider cosine rule.
    Thanks very much! I am a bit stuck on working out the resultant velocity. Could you please help me here?
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    (Original post by old_engineer)
    The new velocity diagram is correct. Now consider that the plane is already 400km due east of Heathrow when this diagram comes into play, as the first hour was wind-free. Prague lies a further 800km due east. Where will the plane be after a further two hours of flying? It will be a further 800km east plus 150km north-east (your vectors times two hours). It’s displacement from Prague is then...
    Thanks very much! I just got confused where you mentioned the further 150km.. Could you please help me here?
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    (Original post by sienna2266)
    Thanks very much! I just got confused where you mentioned the further 150km.. Could you please help me here?
    The displacement of the plane due to the wind is 75kmh^-1 x 2 hours = 150km
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    (Original post by old_engineer)
    The displacement of the plane due to the wind is 75kmh^-1 x 2 hours = 150km
    But isn't this just doing the speed of the wind multiplied by 2 hours? Aren't we meant to consider the resultant velocity in this situation as this is the speed the plane is going as a result of the wind?
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    (Original post by sienna2266)
    But isn't this just doing the speed of the wind multiplied by 2 hours? Aren't we meant to consider the resultant velocity in this situation as this is the speed the plane is going as a result of the wind?
    Part (ii) simply asks how far the plane will be from Prague at the end of 3 hours. My suggestion is that, at the end of 3 hours, the plane will be 3 x 400km due east of Heathrow due to its own motion through the air, plus 2x 75km north-east due to the wind. Net effect is 150km NE of Prague.
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    (Original post by old_engineer)
    Part (ii) simply asks how far the plane will be from Prague at the end of 3 hours. My suggestion is that, at the end of 3 hours, the plane will be 3 x 400km due east of Heathrow due to its own motion through the air, plus 2x 75km north-east due to the wind. Net effect is 150km NE of Prague.
    Thank you! I am struggling to do the diagram for iii) as the wind and the resultant velocity overlap.. Could you please help here?
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    (Original post by RDKGames)
    Work out the resultant velocity.

    The question basically asks in part (i) how long does the normal journey takes in the simplest case with no resistant forces. T=3 as found.

    In part (ii), you are considering the journey of length 3 hours, where for the first hour the plane flies as usual. We are now interested to see where it ends up after 2 hours going at the resultant velocity. So the next step is to determine how much distance the plane covers at speed R going for 2 hours. Also, you wish to determine the bearing of R to see in what direction the plane is offset from the original course.
    Once you have direction and speed of the resultant path, simply consider drawing a displacement diagram between two horizontal points H and P 1200km apart representing the airports, and now draw on the path that the plane takes using the info previously. Label the point it ends up at as A. You wish to determine distance AP. Consider cosine rule.
    Thank you! I wanted to have a go at doing part ii) this way as well.
    I am a bit stuck on working out the resultant velocity. Could you please help me here?
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    (Original post by sienna2266)
    Thank you! I am struggling to do the diagram for iii) as the wind and the resultant velocity overlap.. Could you please help here?
    The direction of the resultant velocity has to be due east. The wind vector points towards NE as before, and it’s magnitude is known. The plane’s velocity vector is the missing side of the triangle (which must be arranged such that plane vector plus wind vector = resultant vector).
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    (Original post by old_engineer)
    The direction of the resultant velocity has to be due east. The wind vector points towards NE as before, and it’s magnitude is known. The plane’s velocity vector is the missing side of the triangle (which must be arranged such that plane vector plus wind vector = resultant vector).
    for iii) isn't the resultant velocity going south west though because from part ii) we know it ended up north east so to get to prague it has to go southwest?
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    (Original post by sienna2266)
    for iii) isn't the resultant velocity going south west though because from part ii) we know it ended up north east so to get to prague it has to go southwest?
    No. The resultant velocity has to be pointing directly towards Prague, i.e due east, as that is where the plane is trying to get to. The plane’s velocity will need to be directed towards somewhere between E and SE to achieve this.
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    (Original post by old_engineer)
    No. The resultant velocity has to be pointing directly towards Prague, i.e due east, as that is where the plane is trying to get to. The plane’s velocity will need to be directed towards somewhere between E and SE to achieve this.
    Ohh so part ii) is not related to part iii)? So for part iii) the plane is not 150km north east from prague to begin with, right?
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    (Original post by sienna2266)
    Ohh so part ii) is not related to part iii)? So for part iii) the plane is not 150km north east from prague to begin with, right?
    I took part (iii) to mean “what course should the pilot set if he actually wants to arrive at the originally planned destination?”

    Other interpretations may be possible, though.
 
 
 
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