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!

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!

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

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