We are tasked with calculating the time it took for the bullet to travel a distance of 23.7 cm (0.237 m) with a final muzzle velocity of 677 m/s, assuming constant acceleration. The bullet starts from rest, so its initial velocity $v_0 = 0$.

We can use the following kinematic equation for constant acceleration:

$v^2 = v_0^2 + 2 a d$

Where:

• $v = 677 \, \text{m/s}$ (final velocity),
• $v_0 = 0 \, \text{m/s}$ (initial velocity),
• $d = 0.237 \, \text{m}$ (distance traveled),
• $a$ is the acceleration.

Step 1: Solve for acceleration

$677^2 = 0^2 + 2 a (0.237)$ $458329 = 2 a (0.237)$ $458329 = 0.474 a$ $a = \frac{458329}{0.474} \approx 967136.71 \, \text{m/s}^2$

Step 2: Find the time taken to leave the barrel

Using another kinematic equation to find the time $t$:

$v = v_0 + a t$

Since $v_0 = 0$:

$t = \frac{v}{a} = \frac{677}{967136.71} \approx 0.0007 \, \text{seconds}$

To convert this into milliseconds:

$t = 0.0007 \, \text{seconds} \times 1000 = 0.7 \, \text{milliseconds}$

Final Answer:

It took approximately 0.7 milliseconds for the bullet to leave the barrel.

Do you need more details or have any questions?

Here are 5 related questions:

1. What is the significance of assuming constant acceleration in this problem?
2. How would air resistance affect the time calculation?
3. What would happen if the muzzle velocity was doubled?
4. How does the length of the barrel influence the bullet's acceleration?
5. Can the same method be used if the bullet had an initial velocity?

Tip: Always double-check unit conversions, especially when switching between meters, centimeters, and seconds.