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Average velocity is a fundamental concept in kinematics and refers to the total displacement divided by the total time taken for a particular trip. In simpler terms, it tells us how fast the position changes on average over a specific duration.
For example, in our exercise, we calculate the average velocity of a car travelling from a traffic light over a span of 0 to 10 seconds. We use the formula:

• Average velocity, \( \bar{v} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} \)

We plug in the values we calculated for the distance at the specified times: \( x(10) = 120 \text{ meters} \) and \( x(0) = 0 \text{ meters} \). The result is an average velocity of 12 m/s.
This value signifies that if the car traveled with a constant velocity, it would move 12 meters forward every second during this period.

Instantaneous velocity is the velocity of an object at a particular moment in time. It differs from average velocity as it focuses on a specific instant rather than an interval.
To find instantaneous velocities, we need to calculate the derivative of the position function, which reveals the rate of change of position at any given time. The formula derived for velocity in the exercise is:

• \( v(t) = 4.80t - 0.36t^2 \)

By substituting different time values into the formula, we compute the car's velocity at specific times:

• At \( t = 0 \), \( v(0) = 0 \text{ m/s} \)
• At \( t = 5 \), \( v(5) = 15 \text{ m/s} \)
• At \( t = 10 \), \( v(10) = 12 \text{ m/s} \)

These results depict how the car's speed changes as time progresses.

The position function is a mathematical representation describing how an object's position changes over time. In our scenario, the car's position function is given as:\[ x(t) = bt^2 - ct^3 \]where the constants \( b \) and \( c \) are provided as 2.40 and 0.120 respectively. This equation tells us the displacement of the car relative to a starting point, in this case, the traffic light.
The position function is crucial because it helps us determine both the average and instantaneous velocities. By evaluating this function at different times, we can find exact positions which are essential for calculating velocities over specified intervals.

In the context of motion, the derivative of a position function gives us the velocity function. Mathematically, a derivative indicates how a function changes as its input changes. It's a core tool in calculus for understanding rates of change.
Taking the derivative of the car's position function, we get the velocity function:

• Start with \( x(t) = bt^2 - ct^3 \)
• Apply differentiation to find \( v(t) = \frac{d}{dt}(bt^2 - ct^3) = 2bt - 3ct^2 \)

This operation allows us to decide the instantaneous rate of position change (or velocity) for any given moment, aiding in the comprehensive understanding of an object's motion over time.

The rest condition in kinematics refers to the state when an object's velocity is zero. This means the object isn't moving at that specific moment. For the car, we determine when it is at rest by setting its velocity function to zero and solving for time \( t \):\[ v(t) = 4.80t - 0.36t^2 = 0 \]
Solving for \( t \), we find two instances where the car's velocity is zero:

• Initially, \( t = 0 \), before the car begins to move.
• Later, \( t = 13.33 \text{ seconds} \), where it stops again.

Understanding the rest condition helps in analyzing motion scenarios where objects may stop and start, an essential aspect of motion analysis.