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The velocity components of an object describe how its velocity is distributed along the coordinate axes. For a squirrel running in a park, these components can be derived from the position vector, which provides the object's location in terms of time. The position vector for our squirrel is given by \[ \vec{r} = [(0.280 \ \text{m/s}) t + (0.0360 \ \text{m/s}^2)t^2] \, \hat{\imath} + [0.0190 \ \text{m/s}^3 t^3] \, \hat{\jmath} \]. We find the velocity components by differentiating each part of the position vector with respect to time.

• To find the x-component of the velocity, \(v_x(t)\), we differentiate the x-component of the position: \[ v_x(t) = \frac{d}{dt}[(0.280)t + (0.0360)t^2] = 0.280 + 0.072t \].
• For the y-component of the velocity, \(v_y(t)\), we differentiate the y-component of the position: \[ v_y(t) = \frac{d}{dt}[0.0190 t^3] = 0.057 t^2 \].

These equations describe how quickly the squirrel's position changes along the x and y axes.

When analyzing motion, especially in two-dimensional space, the position vector is a critical parameter. This vector tells us exactly where an object is at any given time and is represented in terms of i and j unit vectors, corresponding to the x and y axes, respectively.For the squirrel, the position vector is \[ \vec{r} = [(0.280 \ \text{m/s}) t + (0.0360 \ \text{m/s}^2)t^2] \, \hat{\imath} + [0.0190 \ \text{m/s}^3 t^3] \, \hat{\jmath} \]. Here's how it breaks down:

• The term \((0.280)t\) indicates a uniform motion in the x-direction with a velocity of 0.280 m/s, while \((0.0360)t^2\) adds a component of acceleration, increasing the position more rapidly as time progresses.
• The term \(0.0190 t^3\) reflects changing acceleration in the y-direction, implying a more complex motion where velocity increases with the square of the time.

By substituting a specific time value into the position vector, we can find the precise location of the squirrel. For example, at \( t = 5.00 \text{ s} \), this results in a position of approximately 3.31 meters from the initial point.

Determining both the magnitude and direction of velocity gives a full picture of an object's motion. The magnitude indicates how fast the object is moving, while the direction specifies the path it is following relative to an axis or point.To find these for our squirrel at \( t = 5.00 \text{ s} \), we use the velocity components:

• The magnitude of velocity is given by the formula: \[ |v| = \sqrt{v_x(t)^2 + v_y(t)^2} \]. At \(t=5.00 \text{ s}\), substituting in the earlier calculated values of \(v_x(5) = 0.640 \, \text{m/s}\) and \(v_y(5) = 1.425 \, \text{m/s}\), we find \[ |v| \approx 1.56 \, \text{m/s} \].
• To find the direction, or the angle \( \theta \) relative to the positive x-axis, use:\[ \theta = \tan^{-1}\left(\frac{v_y(5)}{v_x(5)}\right) \].This results in \[ \theta \approx 65.56^\circ \], indicating that the velocity is directed above the x-axis.

Understanding these two attributes helps in visualizing the trajectory and behavior of the moving object.