91Ó°ÊÓ

Average velocity is a key concept in vector analysis, representing the total displacement of an object divided by the time interval during which the displacement occurred. Unlike speed, which is a scalar quantity concerned only with magnitude, velocity is a vector quantity that considers both magnitude and direction.

To find average velocity, first determine the total displacement of the object. In this problem, the car travels south, then east, making it necessary to consider both directions. Displacement is a vector, so these directions factor into our calculations:

• For south: The displacement is negative (assigned a negative value for vector purposes) since this is considered the negative y-direction. After 8 seconds, the car's south displacement is calculated as \(d_s = -144.0 \, \text{m}\).
• For east: The displacement is positive, as east is taken as the positive x-direction, resulting in an eastward displacement of \(d_e = 220.0 \, \text{m}\).

Using the Pythagorean theorem, we find the resultant displacement, which accounts for both directions: \(d = \sqrt{d_s^2 + d_e^2} = 263.0 \, \text{m}\).

Finally, the average velocity is this resultant displacement divided by time: \(v_{avg} = \frac{263.0}{8.00} = 32.875 \, \text{m/s}\). The direction is calculated by using angle \(\theta\) with the east direction: \(\theta = \tan^{-1}\left(\frac{|d_s|}{d_e}\right) \approx 33.1^\circ\). Hence, the average velocity indicates a speed of about \(32.9 \, \text{m/s}\) at \(33.1^\circ\) south of east.

Average acceleration gives us an understanding of how an object's velocity changes over a period of time. In vector terms, we look at both changes in speed and direction.

For the car, the initial velocity (\(v_i\)) was \(-18.0 \, \text{m/s}\) due south (which we take as a negative vector in the y-direction) and after time \(t = 8.0 \, \text{s}\), the velocity changed to \(27.5 \, \text{m/s}\) due east (a positive x-direction vector). These velocities translate into vector components: \(-18.0 \, \mathbf{j}\) and \(27.5 \, \mathbf{i}\), respectively.

The change in velocity (\(\Delta v\)) requires subtracting the initial velocity vector from the final velocity vector. When calculated, the average acceleration (\(a_{avg}\)) is \(\langle 3.44, 2.25 \rangle \, \text{m/s}^2\), with its magnitude given by:

• \(\sqrt{3.44^2 + 2.25^2} \approx 4.11 \, \text{m/s}^2\)

The direction of this average acceleration is towards the east-north, illustrating how the car's velocity increased in magnitude and changed direction during the time interval.

To comprehend displacement calculation, it's essential to perceive it as a vector quantity that describes the change in position of an object. It considers both the magnitude and the direction, differentiating it from distance, which is a scalar quantity.

For the car scenario provided, displacement in both the south and east need to be calculated separately and then combined vectorially. The calculations start with knowing:

• Southward movement gives a displacement of \(d_s = -144.0 \, \text{m}\), using \(v \times t\) through the equation \(18.0 \, \text{m/s} \times 8.00 \, \text{s}\).
• Eastward displacement is \(d_e = 220.0 \, \text{m}\).

The resultant displacement vector, which combines these two, is calculated using the Pythagorean theorem:
\(d = \sqrt{d_s^2 + d_e^2} = 263.0 \, \text{m}\).

This resultant vector describes not just how far the car is from its start, but also its changed position in a specific eastward+southward direction.