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Understanding the position vector is essential as it determines the bullet's exact location in space at a certain time. The position vector \( \mathbf{r}=b \hat{\mathbf{x}}+c t \hat{\mathbf{y}}+d t^{2} \hat{\mathbf{z}} \) provides a unique combination of coordinates that change with time. It combines three components:

• \( b \hat{\mathbf{x}} \): Represents the initial position along the x-axis. Its unit must be meters (m), suggesting that \( b \) itself is a distance component.
• \( c t \hat{\mathbf{y}} \): Represents displacement in the y-direction, changing as time progresses. Here, \( c \) acts as the speed component along the y-axis with units of meters per second (m/s).
• \( d t^2 \hat{\mathbf{z}} \): Describes vertical movement affected by time squared. The constant \( d \) reflects an acceleration component in meters per second squared (m/s²).

This vector efficiently maps the bullet's path, accommodating both initial placement and any movements over time. It is paramount in projectile motion as it captures not just initial conditions, but also successive positions.

Velocity and acceleration are derived from the position vector, explaining how the bullet's movement changes over time. Calculating these derivatives clarifies its speed and rate of change.
First, velocity \( \mathbf{v} \) is obtained by differentiating the position vector \( \mathbf{r} \) with respect to time \( t \):\[ \mathbf{v} = \frac{d}{dt} (b \hat{\mathbf{x}} + c t \hat{\mathbf{y}} + d t^2 \hat{\mathbf{z}}) = 0 \hat{\mathbf{x}} + c \hat{\mathbf{y}} + 2d t \hat{\mathbf{z}}. \]

• Along the x-axis, there's no change, so the velocity is zero.
• For the y-axis, velocity remains constant as \( c \), a form of uniform motion.
• The z-axis presents increasing velocity with time due to \( 2d t \), indicating acceleration.

Next, acceleration \( \mathbf{a} \) is the derivative of velocity:\[ \mathbf{a} = \frac{d}{dt} (0 \hat{\mathbf{x}} + c \hat{\mathbf{y}} + 2d t \hat{\mathbf{z}}) = 0 \hat{\mathbf{x}} + 0 \hat{\mathbf{y}} + 2d \hat{\mathbf{z}}. \]

• There's no acceleration in the x and y directions as they are zero.
• The z-direction continues to show a constant acceleration of \( 2d \), consistent with gravitational influence.

This step-by-step differentiation process is crucial to fully grasp changes in velocity and acceleration during projectile motion.

Unit vectors are fundamental in physics, providing a clear means to describe direction without altering magnitude. In this exercise:

• \( \hat{\mathbf{x}} \): Indicates the direction westward, showing the bullet's initial horizontal aim.
• \( \hat{\mathbf{y}} \): Represents the orthogonal horizontal direction, perpendicular to the bullet's initial path.
• \( \hat{\mathbf{z}} \): Signals the vertical direction downward, aligning with gravitational pull as the bullet moves.

Each unit vector (\( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \)) has a magnitude of 1, ensuring consistency in analyzing directions without changing the scale of physical quantities.
They simplify the calculation, allowing for vector representation of positions, velocities, and accelerations without ambiguity. For students and experts alike, understanding unit vectors is a stepping stone to mastering more complex vector scenarios in physics.