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    1. You are running at a speed of 8 m s–1 in a north-easterly direction (i.e. at 45° to both N and E). Find the components of your velocity in an easterly direction, and in a northerly direction, firstly by drawing, and then by calculation using trigonometry.
    Explain why these two components have the same magnitude.

    You use trigonometry to find the magnitude but since it tells you they're the same, you don't have to find it separatley so therefore I can just work is straight from using the SOH CAH TOA right? I'm not too sure on what to do exactly. Someone please help me.

    Also this question, no clue :eek3:

    2. A train is gradually travelling up a long gradient. The speed of the train is 20 m s–1 and the slope makes an angle of 2° with the horizontal. The summit is 200 m above the starting level. How long will it take to reach the summit?
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    (Original post by Whatever210)
    1. You are running at a speed of 8 m s–1 in a north-easterly direction (i.e. at 45° to both N and E). Find the components of your velocity in an easterly direction, and in a northerly direction, firstly by drawing, and then by calculation using trigonometry.
    Explain why these two components have the same magnitude.
    Okay, so first, construct the diagram you're asked for: north and east are perpendicular axes, and the velocity vector is drawn pointing north-east. Easy stuff.

    Then, you can see it's fairly easy to use trigonometry to find the 'length' (ie: velocity) of the horizontal and vertical components (along the north and east axes respectively).

    And they both have the same magnitude because...

    EDIT:
    2. A train is gradually travelling up a long gradient. The speed of the train is 20 m s–1 and the slope makes an angle of 2° with the horizontal. The summit is 200 m above the starting level. How long will it take to reach the summit?
    Similar idea here; draw a diagram (it doesn't need to be to scale, it's just to help visualise the problem). The hill can be represented as a right-angled triangle, the hypotenuse making an angle of 2° with the horizontal. You also know that the height of the hill is 200 metres. You want to find the distance that the train has to travel, ie: the length of the hypotenuse. You can find this using trigonometry...

    Then speed = (distance / time) using this distance should give you the answer.
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    1.
    The drawing should basically be an x-y axis representing north and east, and a line with angle 45 degrees to both axes (that has magnitude 8-ie8ms-1.
    Component in Northerly direction = 8sin(45) (degrees remember)
    Component in Easterly direcstion=8cos(45)
    They have the same magnitude because cos45 = sin45.

    2.
    Distance to travel along the gradient (ie distance up slope to travel) = use sine rule after drawing a triangle.

    Right angled triangle with angles 2 degrees and 88degrees (and 90 - obviously), the vertical side is labelled 200. Use the sine rule to find the length of the hypotheneuse (ie the length needed to travel up the slope ( clue-more than 200).
    Then simply use speed=distance/time
 
 
 
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