91Ó°ÊÓ

In projectile motion, the velocity equation is a fundamental element when analyzing the travel of an object through a medium. Here, the velocity equation is given as \( v = v_0 - kx \), where:

• \( v \) is the velocity at a distance \( x \) within the medium,
• \( v_0 \) is the initial velocity, assigned a value of 900 ft/s in this case,
• \( k \) is a constant that represents the resistance the medium offers, and
• \( x \) is the distance traveled expressed in feet.

When the projectile comes to rest, the velocity \( v \) becomes zero. This situation provides a convenient way to solve for \( k \). By substituting \( v = 0 \) and \( x = \frac{1}{3} \text{ ft} \), this equation becomes a powerful tool for deriving other quantities such as the resistance constant \( k \) or the time of travel. This illustrates how the velocity equation bridges our understanding between an object's speed, the resistance it experiences, and the resulting motion.

Acceleration is the rate of change of velocity with respect to time. In contexts like projectile motion through a resisting medium, the initial acceleration can be found using the derivative of the velocity equation \( v = v_0 - kx \). To determine this, we use:\[ a = \frac{dv}{dt} = -k \cdot \frac{dx}{dt} \]This expression essentially states that acceleration is the rate at which velocity changes, affected directly by the resistance constant \( k \) and the object's velocity \( v \). When the projectile first enters the medium, its velocity is exactly the initial velocity \( v_0 \). Therefore, the initial acceleration is calculated as:\[ a_0 = -k \times v_0 \]Substituting the known values, \( k = 2700 \text{ ft}^{-1} \) and \( v_0 = 900 \text{ ft/s} \), we find:\[ a_0 = -2700 \times 900 = -2430000 \text{ ft/s}^2 \]This negative acceleration indicates the projectile is decelerating due to the medium's resistance, offering insights into how rapidly the object's speed reduces.

The time it takes for a projectile to penetrate a certain distance within a medium is significant in understanding its motion. In this context, to find the time to penetrate 3.9 inches, which is \( \frac{0.325}{1} \text{ ft} \), we separate variables within the velocity equation \( v = 900 - 2700x \) and integrate. This technique allows us to correlate time directly with the given conditions.Setting up the equation:\[ dt = \frac{dx}{900 - 2700x} \]Integrate to find time from \( x = 0 \) to \( x = 0.325 \) while starting from \( t = 0 \) to some time \( t \):\[ \int_{0}^{t} dt = \int_{0}^{0.325} \frac{dx}{900 - 2700x} \]Upon solving, we derive:\[ t = -\frac{1}{2700} [\ln |900 - 2700 \times 0.325| - \ln |900|] \]Hence, the penetration time worked out to approximately 0.000134 seconds. This value, although small, highlights how quickly a high-speed projectile can penetrate a given medium before slowing down considerably.

Unit conversion is critical when solving physics problems to ensure consistency across all measurements. Here, converting distances from inches to feet is a necessary first step. Knowing that 12 inches make one foot allows us to convert easily:

• Convert 4 inches to feet: \( x = \frac{4}{12} = \frac{1}{3} \text{ ft} \)
• Convert 3.9 inches for the time calculation: \( \frac{3.9}{12} = 0.325 \text{ ft} \)

These conversions ensure that all numbers can be directly used with velocities and accelerative forces expressed in feet and seconds. Units impact calculations significantly, making it crucial to standardize them before proceeding. Not only does this simplify mathematical operations, but it also helps avoid errors and makes interpretations easier, especially when cross-referencing different quantities.