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When determining the average velocity, we are essentially looking for the ratio of the total displacement to the total time during a specific time period. In the context of the exercise, the average velocity provides a simple measure of how fast an object is moving over the entire interval.
For a car traveling from a stop at a traffic light, we calculate the displacement by evaluating the position function at two different times, specifically at the beginning and the end of the interval, then dividing by the time elapsed. Here, the position function given is \[ x(t) = 2.4t^2 - 0.120t^3 \].To compute the average velocity from \(t = 0\) to \(t = 10\) seconds, we find the displacement as \(x(10) - x(0)\), giving us the formula:\[ v_{avg} = \frac{x(10) - x(0)}{10 - 0} \].By substituting the values, the computed average velocity is \(12 \text{ m/s}\).
Remember:

• Average velocity factors in the entire trip.
• It doesn't account for variations in speed that might occur in-between.

Instantaneous velocity, in contrast to average velocity, tells us how fast a car is moving at a particular moment in time. It is the derivative of the position function with respect to time, representing the rate of change of displacement.
For the position function:\[ x(t) = 2.4t^2 - 0.120t^3 \],the velocity function is the derivative:\[ v(t) = \frac{d}{dt}(2.4t^2 - 0.120t^3) = 4.8t - 0.36t^2 \].By substituting specific times into this velocity function, we can determine the car's speed at that moment:

• At \(t = 0\), \(v(0) = 0\, \text{m/s}\), meaning the car is initially at rest.
• At \(t = 5\), \(v(5) = 12\, \text{m/s}\).
• At \(t = 10\), \(v(10) = 0\, \text{m/s}\), indicating the car is momentarily at rest again.

Instantaneous velocity provides useful insight on how the car's speed changes over time.

Displacement refers to the change in position of the car, indicating how far the car has moved from its starting point and in what direction. It can be calculated by evaluating the difference in position at two separate times. Displacement is a vector quantity and considers both magnitude and direction.
In the exercise, the position of the car as a function of time is given by:\[ x(t) = 2.4t^2 - 0.120t^3 \].To find the displacement between time intervals, substitute the times into this equation:

• At \(t = 0\): The position is \(x(0) = 0\, \text{m}\).
• At \(t = 10\): The position is \(x(10) = 240\, \text{m} - 120\, \text{m} = 120\, \text{m}\).

So, the displacement over this interval is \(120\, \text{m}\). Remember:

• Displacement is straight-line distance from start to end.
• It differs from distance which is the total path traveled.

Acceleration describes how quickly the velocity of an object is changing. It's the rate of change of velocity with respect to time and can be either positive or negative.
In the exercise, to find the car's acceleration, we need to take the derivative of the velocity function:\[ v(t) = 4.8t - 0.36t^2 \].Hence, the acceleration \(a(t)\) is:\[ a(t) = \frac{d}{dt}(4.8t - 0.36t^2) = 4.8 - 0.72t \].This tells us how acceleration varies over time. For this car:

• At \(t = 0\), \( a(0) = 4.8\, \text{m/s}^2\), meaning it starts with a strong acceleration.
• As \(t\) increases, the acceleration reduces because \(-0.72t\) term becomes more significant.

The car experiences different acceleration values at different times, influencing its velocity and position. It shows slowing down when the acceleration becomes negative.

Velocity-time graphs provide a visual representation of how an object's velocity changes over time. On these graphs, the time is on the horizontal axis and velocity is on the vertical axis.
Analyzing such graphs gives insights into the object’s motion patterns:

• When the graph is a horizontal line, the velocity is constant.
• If the line slopes upwards, the velocity is increasing (accelerating).
• If the line slopes downwards, the velocity is decreasing (decelerating).

For our car example, the velocity function is:\[ v(t) = 4.8t - 0.36t^2 \].By plotting this, we can see how the velocity changes:

• It starts at zero, rises, and then falls back to zero.
• The highest point on the graph signifies the maximum velocity.

The graph helps in understanding when the car speeds up and when it starts to slow down, providing a comprehensive picture of its overall motion.