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    This is in for tomorrow and I'm completely clueless @[email protected]

    A particle travelling in a straight line at 15m/s is brought to rest with constant deceleration in a distance of 22.5m. Show that the deceleration takes 3 seconds.

    Any hints?
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    ooh dont u do s=1/2(v-u)t

    so 22.5 = 0.5 * t * (0-15)
    22.5=-7.5t
    t=-3 or 3 seconds
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    s=0.5(u+v)t
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    Oh those equations.. x_x
    Thanks!
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    Another question I'm stuck on..

    A ball A is thrown vertically upwards at 25m/s from a point P. Three seconds later a second ball B is also thrown vertically upwards from the point P at 25m/s. Taking the acceleration due to gravity to be 10m/s, calculate:
    i.) the time for which ball A has been in motion when the balls meet
    ii.) the height above P at which A and B meet
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    (Original post by smexyace)
    Another question I'm stuck on..

    A ball A is thrown vertically upwards at 25m/s from a point P. Three seconds later a second ball B is also thrown vertically upwards from the point P at 25m/s. Taking the acceleration due to gravity to be 10m/s, calculate:
    i.) the time for which ball A has been in motion when the balls meet
    ii.) the height above P at which A and B meet
    Form a simultaneous equation, I may be wrong. I vaguely remember doing this question.
 
 
 
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Updated: November 10, 2008

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