91Ó°ÊÓ

The concept of motion in the negative x-direction implies that both the velocity and acceleration of an object are oriented towards the negative side of a coordinate system's x-axis. In simpler terms, if you picture a number line with zero at the center, moving in the negative x-direction means moving left towards negative values.
For this exercise, the car started and ended with velocities of -25 m/s and -50 m/s respectively, both in the negative x-direction. This is crucial because it implies that all calculations, such as those for velocity change and acceleration, need to consider the direction as indicated by the negative sign.
When direction is specified, it helps us understand not only how fast something is moving, but also where it's headed, which is essential for solving problems involving velocity and acceleration.

Velocity change refers to the difference between the initial and final velocities of an object. It's a vector quantity, meaning it has both magnitude and direction. To find the change in velocity, subtract the initial velocity from the final velocity.

• In our exercise, the initial velocity ( v_i) was -25 m/s and the final velocity ( v_f) was -50 m/s. This yields a velocity change ( \Delta v) calculated as follows:
  \[ \Delta v = v_f - v_i = -50 \text{ m/s} - (-25 \text{ m/s}) = -25 \text{ m/s}\]

The negative sign indicates that the change is also in the negative x-direction, meaning the car is speeding up in that direction. Understanding this change is vital because it directly affects how we calculate other dynamics, like acceleration, of the car's movement.

When discussing the distance traveled with constant acceleration, we consider both the initial velocity and the acceleration. The distance can be computed using the formula:
\[s = v_i t + \frac{1}{2} a_{avg} t^2\]
In this formula:

• v_i represents the initial velocity, which in this example was -25 m/s.
• t stands for the time period, which was 5 seconds in our case.
• a_{avg} denotes the average acceleration, calculated as -5 m/s².

Applying these values:
\[ s = (-25 \text{ m/s})(5 \text{ s}) + \frac{1}{2} (-5 \text{ m/s}^2)(5 \text{ s})^2\]
which simplifies to:
\[ s = -125 \text{ m} - 62.5 \text{ m} = -187.5 \text{ m}\]
The negative value of s confirms the car moved 187.5 meters further in the negative x-direction, aligning with the initial velocity direction.

Constant acceleration implies that the rate of change of velocity is steady over time. This means, in each second of travel, the velocity changes by a consistent amount. For this exercise, the car had a constant acceleration of -5 m/s².
To determine this, the formula for average acceleration is used:

• \[a_{avg} = \frac{\Delta v}{\Delta t}\], where \Delta v is the change in velocity and \Delta t is the time period.

Substituting the values:
\[ a_{avg} = \frac{-25 \text{ m/s}}{5 \text{ s}} = -5 \text{ m/s}^2\]
The negative sign indicates the acceleration direction is the same as the motion direction (negative x-direction). Constant acceleration simplifies calculations and predictions about motion, as it ensures that velocity increases or decreases uniformly, making complex kinematic problems more manageable.