The barrel of a rifle has a length of 0.976 m. A

bullet leaves the muzzle of a rifle with a speed
of 561 m/s.
What is the acceleration of the bullet while
in the barrel? A bullet in a rifle barrel does
not have constant acceleration, but constant
acceleration is to be assumed for this problem.
Answer in units of m/s
2

Well, if you assume constant acceleration, then we can use the equation of motion:

v^2 = u^2 + 2as

where v is the final velocity (561 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the displacement (0.976 m).

Since the initial velocity is zero, the equation simplifies to:

v^2 = 2as

Plugging in the values we have:

(561 m/s)^2 = 2a(0.976 m)

a = (561 m/s)^2 / (2 * 0.976 m)

The acceleration of the bullet while in the barrel is approximately 164,628.48 m/s^2.

But remember, this is assuming constant acceleration, which is not necessarily the case for a bullet in a rifle barrel. So take this answer with a grain of gunpowder!

To find the acceleration of the bullet while in the barrel, we can use the equation for constant acceleration:

v^2 = u^2 + 2as

Where:
- v is the final velocity (561 m/s)
- u is the initial velocity (0 m/s since the bullet starts from rest in the barrel)
- a is the acceleration of the bullet while in the barrel
- s is the distance traveled in the barrel (0.976 m)

Rearranging the equation, we have:

a = (v^2 - u^2) / 2s

Substituting the given values, we get:

a = (561^2 - 0^2) / (2 * 0.976)

a = 314721 / 1.952

Calculating the value, the acceleration of the bullet while in the barrel is approximately:

a = 161089.8459 m/s^2 (rounded to the nearest thousandth)

Therefore, the acceleration of the bullet while in the barrel is 161089.8459 m/s^2.

To find the acceleration of the bullet while in the barrel, we need to determine the time it takes for the bullet to travel through the barrel.

We can use the equation of motion:

v = u + at

Where:
v = final velocity (561 m/s)
u = initial velocity (0 m/s)
a = acceleration (to be determined)
t = time

Since the bullet starts from rest (u = 0 m/s), the equation simplifies to:

v = at

Now, rearrange the equation to solve for the acceleration (a):

a = v / t

We know the final velocity (v = 561 m/s), so we need to find the time (t).

To find the time, we can use another equation of motion:

s = ut + (1/2)at^2

Where:
s = distance (length of the barrel = 0.976 m)
u = initial velocity (0 m/s)
t = time
a = acceleration (to be determined)

Rearrange this equation to solve for time (t):

0.976 = (1/2)at^2

Rearrange again to isolate t:

t^2 = (2 * 0.976) / a

t = sqrt((2 * 0.976) / a)

Now, substitute the expression for time (t) back into the acceleration equation:

a = v / t
a = 561 / sqrt((2 * 0.976) / a)

Simplify the equation:

a^2 = (561^2 * (2 * 0.976)) / a

Multiply both sides by a and simplify:

a^3 = (561^2 * 2 * 0.976)

Now, take the cube root of both sides to solve for acceleration (a):

a = (561^2 * 2 * 0.976)^(1/3)

Calculating the expression gives:

a ≈ 3,897.45 m/s^2

Therefore, the acceleration of the bullet while in the barrel is approximately 3,897.45 m/s^2.

V^2 = Vo^2 + 2a*d

V = 561 m/s
Vo = 0
d = 0.976 m.
a = ?