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.

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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!

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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!

Quote:

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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!

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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

Quote:

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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!

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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