91Ó°ÊÓ

When exploring the motion of an object, particularly vertical motion as in the exercise given, we come across the concept of average velocity. Simply put, it is the consistent speed at which an object would need to travel to cover a certain distance in a specific time period. Think of it as the 'overall' speed of the object when you don't care about the ups and downs of its actual path.

In mathematical terms, average velocity (\(v_{avg}\)) is calculated by finding the difference in displacement over a certain interval and dividing it by the time elapsed during that interval. In the exercise provided, the function representing the height of the object over time, given as \[s(t)=-4.9t^2+300\], provides essential information to calculate average velocity.

To calculate the average velocity during the first 3 seconds, one would take the objects position at 3 seconds (\(s(3)\)), subtract the initial position \(s(0)\), which gives the overall displacement, and divide by the time interval, 3 seconds in this case. This gives a straightforward measure of how quickly the object is moving 'on average' during this time frame.

Diving into the concept of instantaneous velocity, we find ourselves peering into a more 'zoomed-in' view of an object's motion compared to average velocity. Instantaneous velocity is essentially the velocity of an object at a specific, singular moment in time. Imagine pausing a video of a falling object right at the 1-second mark and being able to tell exactly how fast it was moving at that split second — that's what instantaneous velocity represents.

This concept is critical in calculus, as it relates to the derivative of the displacement function with respect to time. In the given exercise, the displacement function \(s(t) = -4.9t^2 + 300\) can be differentiated to find the function for instantaneous velocity, represented by \(s'(t)\).

The instantaneous velocity at any time \(t\) can be calculated by taking the derivative of the displacement function, which, in this case, yields \(s'(t) = -9.8t\). This derivative function tells us the exact velocity of the object at any point in time, which is why it can vary from the average velocity, except at specific points where the Mean Value Theorem assures us they must be equal.

The process of differentiating a displacement function is essential in determining the instantaneous rates of change, such as velocity. In the world of calculus, differentiation allows us to take a function that describes something's position relative to time and determine the rate at which that position is changing at any given moment.

For the exercise in question, the displacement function \(s(t) = -4.9t^2 + 300\) describes the height of an object over time. When we differentiate this function with respect to time, we are essentially obtaining the formula for the object's velocity - in this case, how quickly the object is moving upwards or downwards at any point in time.

\begin{align*}s(t) &= -4.9t^2 + 300 \ s'(t) &= \frac{d}{dt} (-4.9t^2 + 300) \ &= -9.8t\text{.}end{align*}This differentiation gives us \(s'(t)\), which can then be set equal to the average velocity to find the specific time at which an object's instantaneous velocity is equal to its average velocity, satisfying the conditions of the Mean Value Theorem.