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Size:  109.0 KB for a) why is the angle not 56.3 but instead 90+56? I thought this is the angle we are looking for Attachment 511575511577 and for b) why isn't it just (2i-3j)t? In the mark scheme acceleration is included as well but how could that fit into the equation?? Thank you.
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    (Original post by coconut64)
    Name:  Screenshot_2016-03-09-15-32-37.png
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Size:  109.0 KB for a) why is the angle not 56.3 but instead 90+56? I thought this is the angle we are looking for Attachment 511575511577 and for b) why isn't it just (2i-3j)t? In the mark scheme acceleration is included as well but how could that fit into the equation?? Thank you.
    The angle you've found is the angle between \mathbf{i} and the direction of motion, not the angle requested.
    As for part b) think suvat:
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    So \mathbf{v}=\mathbf{v_0}+t\mathbf  {a}.
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    (Original post by joostan)
    The angle you've found is the angle between \mathbf{i} and the direction of motion, not the angle requested.
    As for part b) think suvat:
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    So \mathbf{v}=\mathbf{v_0}+t\mathbf  {a}.
    Oh I get it. For your equation is that the v=u+at equation? Is v0 u? Cheers.
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    (Original post by coconut64)
    Oh I get it. For your equation is that the v=u+at equation? Is v0 u? Cheers.
    Yes, \mathbf{v_0} is an initial velocity.
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    (Original post by joostan)
    Yes, \mathbf{v_0} is an initial velocity.
    Now this makes sense. So this euqation is used when you only know the intial v and a ?
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    (Original post by coconut64)
    Now this makes sense. So this euqation is used when you only know the intial v and a ?
    This is equivalent to component-wise suvat and applies for constant acceleration.
    You can use this exactly as you used the single dimensional equivalent v=u+at.
 
 
 
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