Problem on One-Dimentional Motion: Bullet going through a board.

[redredredred]

An indestructible bullet 2.00cm long is fired straight through a board that is 10.0cm thick. The bullet strikes the board with a speed of 420 m/s and emerges with a speed of 280m/s.
(a) What is the average acceleration of the bullet through the board?
(b) What is the total time that the bullet is in contact with the board?
(c) What thickness of board (calculated to 0.1cm) would it take to stop the bullet, assuming that the acceleration through all boards is the same?

We can use these equations (v₀=initial velocity, D=displacement):

v = v₀ + at
D = v₀t + (1/2)at²
v2 = v₀² + 2aD
D = (1/2)(V₀+V)t
D = vt - (1/2)at²



I have found that the answer to part (a) is equal to -490,000m/s²,
but I cannot figure out how to do parts (b) and (c).
Please help?!

 

[anathema]

(a) The average acceleration of the bullet through the board can be calculated using the equation: a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time. Plugging in the given values, we get: a = (280 m/s - 420 m/s)/t = -140 m/s2/t.

(b) To find the total time that the bullet is in contact with the board, we can use the equation: D = v0t + (1/2)at2, where D is the displacement, v0 is the initial velocity, a is the acceleration, and t is the time. We know that the displacement is equal to the thickness of the board, which is 10.0cm = 0.1m. We also know that the initial velocity is 420 m/s and the acceleration is -140m/s2. Plugging in these values, we get: 0.1m = 420 m/s * t + (1/2)(-140 m/s2)t2. This is a quadratic equation that can be solved using the quadratic formula. The solutions are t = 0.001 seconds and t = 0.002 seconds. Since we are interested in the total time that the bullet is in contact with the board, we take the larger value, which is t = 0.002 seconds.

(c) To find the thickness of the board that would stop the bullet, we can use the equation: D = vt - (1/2)at2, where D is the thickness of the board, v is the final velocity, a is the acceleration, and t is the time. We know that the final velocity is 0 m/s, the acceleration is -140 m/s2, and the time is 0.002 seconds (as calculated in part (b)). Plugging in these values, we get: D = (0 m/s)(0.002 s) - (1/2)(-140 m/s2)(0.002 s)2 = 0.00028 m = 0.028 cm. Therefore, the thickness of the board that would stop the bullet is 0.028 cm.

 

[baba89]

(a) The average acceleration of the bullet through the board can be found using the equation:

a = (vf - vi) / t

Where vf is the final velocity (280 m/s) and vi is the initial velocity (420 m/s). We also know that the displacement (D) is equal to the thickness of the board (10 cm or 0.1 m). Therefore, we can rewrite the equation as:

a = (280 m/s - 420 m/s) / t

To find t, we need to use the equation:

D = vt - (1/2)at^2

Substituting the values we know, we get:

0.1 m = (280 m/s)t - (1/2)(-490,000 m/s^2)t^2

Solving for t, we get t = 0.000408 s.

Now, we can go back to the first equation and substitute the value of t to find the average acceleration:

a = (280 m/s - 420 m/s) / 0.000408 s

Therefore, the average acceleration of the bullet through the board is equal to -480,392.16 m/s^2.

(b) To find the total time that the bullet is in contact with the board, we can use the equation:

D = (1/2)(V0+V)t

Substituting the values we know, we get:

0.1 m = (280 m/s + 420 m/s)t / 2

Solving for t, we get t = 0.000238 s.

Therefore, the total time that the bullet is in contact with the board is 0.000238 seconds.

(c) To find the thickness of the board that would stop the bullet, we can use the same equation as in part (a):

a = (vf - vi) / t

Now, we know that the final velocity (vf) is 0 m/s, since the bullet would stop when it hits the board. Substituting the values we know, we get:

0 m/s = (420 m/s - 280 m/s) / t

Solving for t, we get t = 0.000476 s.

Now, we can use this value of t in the equation for displacement to find the thickness of the board (D):

D = (1/2)(V0+V)t

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