91Ó°ÊÓ

The position vector \( \vec{r} \) describes the location of a particle in a plane at any given time \( t \). In our exercise, the position vector is provided as a simple function of \( t \), expressed in terms of the unit vectors \( \hat{i} \) and \( \hat{j} \) denoting directions along the \( x \) and \( y \) axes respectively:
\[ \vec{r} = (2.00 t^3 - 5.00 t) \hat{i} + (6.00 - 7.00 t^4) \hat{j} \]
The components of this formula reveal how movement is achieved in each direction over time. The expression \( 2.00 t^3 - 5.00 t \) identifies the particle's motion along the \( x \)-axis, whereas \( 6.00 - 7.00 t^4 \) reveals motion along the \( y \)-axis.

To find the specific position of the particle at \( t = 2.00 \) seconds, we substitute \( t \) with this value and simplify our equations:

• \( x \)-component: \( 2.00(2.00)^3 - 5.00(2.00) = 6.00 \) meters
• \( y \)-component: \( 6.00 - 7.00(2.00)^4 = -106.00 \) meters

Thus, the position vector is \( \vec{r} = 6.00 \hat{i} - 106.00 \hat{j} \), showing a meaningful placement within the \( xy \) plane.

The velocity vector \( \vec{v} \) is crucial for understanding the movement dynamics of a particle as it encapsulates the rate of change of its position. Mathematically, it is obtained by differentiating the position vector with respect to time.
The general expression for velocity in our case follows as:
\[ \vec{v} = \frac{d}{dt}\left((2.00 t^3 - 5.00 t) \hat{i} + (6.00 - 7.00 t^4) \hat{j}\right) \]
This differentiation results in:
\[ \vec{v} = (6.00 t^2 - 5.00) \hat{i} + (-28.00 t^3) \hat{j} \]
To pinpoint the velocity at \( t = 2.00 \) seconds, substitute \( t \) with this value and solve:

• \( v_i \) (\( \hat{i} \)-component): \( 6.00(4) - 5.00 = 19.00 \) m/s
• \( v_j \) (\( \hat{j} \)-component): \( -28.00(8) = -224.00 \) m/s

Therefore, the velocity vector is \( \vec{v} = 19.00 \hat{i} - 224.00 \hat{j} \), detailing the particle's momentum along both axes at this specific time.

The acceleration vector \( \vec{a} \) elucidates the change in velocity over time and is derived by taking the derivative of the velocity vector. This tells us how quickly the velocity itself is changing, providing insight into factors like speeding up or slowing down.

For our problem, the acceleration vector is obtained from:
\[ \vec{a} = \frac{d}{dt}\left((6.00 t^2 - 5.00) \hat{i} - 28.00 t^3 \hat{j}\right) \]
Solving this, we find:
\[ \vec{a} = 12.00 t \hat{i} - 84.00 t^2 \hat{j} \]
To find acceleration at \( t = 2.00 \) seconds, substitute the value into the equation:

• \( a_i \) (\( \hat{i} \)-component): \( 12.00(2) = 24.00 \) m/s²
• \( a_j \) (\( \hat{j} \)-component): \( -84.00(4) = -336.00 \) m/s²

Consequently, the acceleration vector becomes \( \vec{a} = 24.00 \hat{i} - 336.00 \hat{j} \), illustrating how forces act to modify movement in both the \( x \) and \( y \) directions.

The tangent angle to a path is a fundamental idea that links geometry with kinematics. It illustrates the orientation of a particle's path by representing the angle between the path tangent and a reference axis, often the positive \( x \)-axis.

In our exercise, this tangent angle \( \theta \) at \( t = 2.00 \) seconds can be computed using the components of the velocity vector:\[ \tan(\theta) = \frac{v_j}{v_i} \]
Plugging in the values, we have:
\[ \tan(\theta) = \frac{-224.00}{19.00} \]
Therefore, \( \theta = \arctan\left(\frac{-224.00}{19.00}\right) \), calculated approximately as \( -85.17^\circ \).
This angle suggests that, at this specific moment, the path of the particle is largely directed along the negative \( y \)-axis, just slightly off from being directly vertical. Such understanding helps visualize motion and anticipate subsequent movement directions.