To find the acceleration of the tranquilizer projectile, we can use one of the equations of motion, specifically the equation that relates acceleration to initial velocity, final velocity, and distance traveled:

$v^2 = u^2 + 2as$

Where:

• $v$ is the final velocity ($521$ m/s),
• $u$ is the initial velocity ($0$ m/s, since it starts from rest),
• $a$ is the acceleration (which we want to find),
• $s$ is the distance traveled ($0.740$ meters).

Plugging in the given values:

$521^2 = 0^2 + 2 \cdot a \cdot 0.740$

$271441 = 1.48a$

Now, solve for $a$:

$a = \frac{271441}{1.48}$

$a \approx 183000 \, \text{m/s}^2$

The acceleration of the projectile is approximately $183,000$ m/s$^2$.

Would you like more details, or do you have any questions?

Here are 5 related questions that expand on this information:

1. What would be the time taken by the projectile to travel through the barrel if the acceleration is $183,000$ m/s$^2$?
2. How would the acceleration change if the barrel length were doubled?
3. What impact does barrel length have on the final speed of a projectile assuming uniform acceleration?
4. What would be the final velocity if the projectile accelerates at the same rate but starts from an initial velocity of $100$ m/s?
5. How does non-uniform acceleration affect the calculations of final speed and acceleration?

Tip: Always use consistent units (like meters and seconds) when working with physics equations to avoid errors in calculations.