ahmadu university zaria nigeria

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A particles start from rest and acceleration passing 2 point (A and B)which are 100cm apart in 15seconds if the velocity of the particle at B is three times it velocity at A,find i)the velocity of the particles at A. ii)the accelation. iii) the distance of A from the starting piont.

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    The language is a bit unclear, but it appears that it takes 15 seconds to travel the 100 cm from A to B. If so,

    since velocity
    v = at

    3at = a(t+15)
    t = 15/2 s

    Since distance
    s = 1/2 at^2,

    100 + 1/2 at^2 = 1/2 a(t+15)^2
    a = 4/9 cm/s^2

    So, we have
    v@A = (4/9)(15/2) = 10/3
    v@B = (4/9)(45/2) = 10
    s@A = 1/2 (4/9)(15/2)^2 = 25/2
    s@B = 1/2 (4/9)(45/2)^2 = 225/2
    Note that 225/2 = 25/2 + 100 as required

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