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An object dropped from a cliff falls with a constant acceleration of 10m/s2. Find its speed 2s after it was dropped?
Solution:Given ,accelaration =10m/s
time=2s
speed=?
therfore,speed=distance travelled\total time taken
s=10\2=5(ans) . is it ok.

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  1. Not really.
    For an object which accelerates from rest, the variation of speed with time is given by:
    final velocity=initial velocity + acceleration * time
    = 0 + 10 m/sยฒ * 2 s
    =20 m/s
    (note how the units cancel to give m/s)

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  2. We will apply v=u+at . 0+2ร—10= 20m/s

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  7. As from the kinematics equation
    V=u+at
    So, u = 0
    a = 10m/s2
    t = 2 s
    V = 0 + 10 * 2 = 20 m/s

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  8. Given,
    a = 2 m/s^2
    t = 10 s
    u =0(Initial speed)
    v = ? (Final speed)

    By the equation v = u + at
    v = 0 + 2*10
    = 0 + 20
    = 20 m/s
    Therefore the speed 2 sec after it was dropped is 20 m/s

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