91Ó°ÊÓ

Average velocity provides a simple measurement of how fast an object is moving over a specific period of time. It is defined as the total displacement divided by the total time taken for that displacement. In the context of projectile motion, like a ball thrown into the air, we can calculate the average velocity over a time interval \[a, b\] using the formula:
\begin{align*}v_{\text{avg}} = \frac{s(b) - s(a)}{b - a}\end{align*}
where \(s(a)\) and \(s(b)\) are the heights of the projectile at the beginning and end of the interval, respectively. Displacement is the key here: it's the change in position, not the total distance traveled. In the given exercise, the average velocity calculations over various time intervals \([2,3], [2,2.5], and [2,2.1]\) help students grasp the concept that the average velocity can change with different intervals, even when the start point remains consistent. This reinforces understanding the effect of time and displacement on average velocity.

Unlike average velocity, instantaneous velocity refers to how fast an object is moving at a very specific point in time. To find this, we often turn to calculus, utilizing the derivative of a function that describes the object's position in terms of time. The derivative gives us the rate of change of position (or height, in the case of a projectile) with respect to time, which is the instantaneous velocity.
For the problem at hand with the equation \(s(t) = 128t - 16t^{2}\), the derivative tells us the instantaneous velocity is \(v(t) = 128 - 32t\). By evaluating this derivative at a given time, such as \(t=2\) or \(t=5\), we can determine the velocity of the ball at precisely those moments. This demonstrates the instantaneous nature of the concept and how it differs from average velocity, which spans a time interval.

The derivative of the height function is central to understanding motion in physics, particularly for projectiles. In calculus terms, the derivative represents the instantaneous rate of change. Here, by differentiating the height function \(s(t) = 128t - 16t^{2}\), we determine how the height of the ball changes at any given moment in time.
The resulting function from differentiating \(s(t)\), which is \(v(t)=128 - 32t\), is essentially the velocity function. It shows that for every second that passes, there is an instantaneous change in the height that can be pinpointed using this function. Applying this to the scenario at hand can help decipher not just the velocity at any point, but also whether the ball is rising or falling by examining the sign of the velocity, which is an invaluable insight when analyzing the motion of projectiles.

The motion of a projectile, such as a ball launched into the air, is a classic problem in physics that illustrates the effects of gravity and initial velocity. The equation \(s(t) = 128t - 16t^{2}\) provides a model of vertical motion, where the term \(-16t^{2}\) describes the downward pull of gravity, and \(128t\) represents the upwards thrust from the initial toss.
Understanding this projectile's motion requires analyzing how its height changes over time, calculating velocities, and ultimately determining when it makes its journey back to the ground. The moment the height function equals zero is the time of impact. Students learn about the symmetry of the projectile's path and how gravity ensures what goes up must come down, all illustrated mathematically by the height function and its derivative, the velocity.