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    " the time in which any space is traversed by a body starting from rest and uniformly accelerated is equal to the time in which that same space would be traversed by the same body moving at a uniform speed whose value is the mean of the highest speed and the speed just before acceleration began"
    I dont quite get the underlined bit , how it relates to the starting statement.
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    (Original post by rbnphlp)
    " the time in which any space is traversed by a body starting from rest and uniformly accelerated is equal to the time in which that same space would be traversed by the same body moving at a uniform speed whose value is the mean of the highest speed and the speed just before acceleration began"
    I dont quite get the underlined bit , how it relates to the starting statement.
    It's speed just before acceleration began, in this case is 0 - its just setting the initial conditions correctly.

    At t=0, acceleration has just started so the body isn't at rest anymore so we're taking the very moment just before the acceleration was applied, when the body was at rest.

    An example;

    A body starts from rest and has constant acceleration b - it reaches a top velocity of a as it travels a set distance - the time taken is t.

    Another body moves with constant velocity (a-0)/2 and takes exactly t to travel the same distance.

    That help at all?
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    If you draw a speed time graph then one will be a triangle and one will be a square and the areas will be the same (when the condition that the uniform speed is the mean for the accelerated one) - might help to visualise it
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    (Original post by kexy)
    It's speed just before acceleration began, in this case is 0 - its just setting the initial conditions correctly.

    At t=0, acceleration has just started so the body isn't at rest anymore so we're taking the very moment just before the acceleration was applied, when the body was at rest.

    An example;

    A body starts from rest and has constant acceleration b - it reaches a top velocity of a as it travels a set distance - the time taken is t.

    Another body moves with constant velocity (a-0)/2 and takes exactly t to travel the same distance.

    That help at all?
    makes sense thankyou
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    (Original post by MC REN)
    If you draw a speed time graph then one will be a triangle and one will be a square and the areas will be the same (when the condition that the uniform speed is the mean for the accelerated one) - might help to visualise it
    are you sure its a square , I thought it would be a rectangle:confused:
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    (Original post by rbnphlp)
    are you sure its a square , I thought it would be a rectangle:confused:
    oh yer, well it would be a rectangle of course

    although in my defence you could just choose the axes scales to be whatever you want and get a 'square' shape
 
 
 
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