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    xgreeneyes98x's Avatar
    xgreeneyes98x Posts: 2, Reputation: 1
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    #1

    Oct 10, 2007, 03:32 PM
    Kinematics- Acceleration
    A bullet traveling 220 m/s strikes a tree and penetrates 4.33 cm before stopping. Find the acceleration of the bullet and the time it takes to stop.

    HELP A.S.A.P!
    BobJ's Avatar
    BobJ Posts: 4, Reputation: 1
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    #2

    Oct 10, 2007, 05:47 PM
    Quote Originally Posted by xgreeneyes98x
    A bullet traveling 220 m/s strikes a tree and penetrates 4.33 cm before stopping. Find the acceleration of the bullet and the time it takes to stop.

    HELP A.S.A.P!
    Solve the two equations for acceleration, a:
    x=1/2 a t^2 or a = 2x/t^2
    v=at or a = v/t

    So 2x/t^2 = v/t
    t=2x/v = 2*.0433 m / 220 m/s
    t = .00039 s

    The acceleration is a = v/t = 220 m/s /.00039 s or 560000 m/s^2, --->57,000g's!
    terryg752's Avatar
    terryg752 Posts: 197, Reputation: 4
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    #3

    Oct 11, 2007, 03:56 AM
    Quote Originally Posted by xgreeneyes98x
    A bullet traveling 220 m/s strikes a tree and penetrates 4.33 cm before stopping. Find the acceleration of the bullet and the time it takes to stop.

    HELP A.S.A.P!
    u = initial velocity = 220, v = final velocity = 0
    s = distance = .0433 m, acceleration = a

    Formula: v^2 -u^2 = 2as

    Hence a = 558891.45 m/sec^2
    pawel's Avatar
    pawel Posts: 12, Reputation: 1
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    #4

    Oct 15, 2007, 07:05 PM
    Quote Originally Posted by xgreeneyes98x
    A bullet traveling 220 m/s strikes a tree and penetrates 4.33 cm before stopping. Find the acceleration of the bullet and the time it takes to stop.

    HELP A.S.A.P!
    Since the bullet has stopped. So we can conclude that bullet must be retarted
    Now as given in the question,
    U=220m/s ,V(final velocity)=0 ,S=distance=0.0433m
    Use third equation of motion i.e v^2-u^2=2as
    putting the above values we get retardation=558891.455m/s^2
    or we can say that acceleration = -558891.455m/s^2
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