Question

In Exercises 101 and \(102,\) use the position function \(s(t)=-16 t^{2}+500\) , which gives the height (in feet) of an object that has fallen for \(t\) seconds from a height of 500 feet. The velocity at time \(t=a\) seconds is given by \(\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\) A construction worker drops a full paint can from a height of 500 feet. How fast will the paint can be falling after 2 seconds?

Short answer

After 2 seconds, the paint can will be falling at a speed of 64 feet per second.

Step-by-step solution

Find the derivative of s(t)

The derivative of a function represents the rate of change of the function at any given point. In the case of position function, the derivative, denoted as s'(t) or ds/dt, gives the velocity of the function. \n\nDifferentiating \(s(t)=-16 t^{2}+500\) will therefore give us the velocity at any time t. Applying the power rule ( which states that the derivative of \(t^n\) is \(n* t^{n-1}\)), when \(n=2\), we have:\n\n \(-16* 2 *t^{2-1} = -32t\). So the derivative of \(s(t)\) is \(-32t\).

Evaluate the derivative at t=2 seconds

Now, to determine the velocity of the can at time t = 2 seconds, substitute 2 into the derivative function. That is, evaluate s'(2).\n\n The expression equals to \(-32 * 2 = -64\). The negative sign indicates that the velocity is downwards, consistent with the direction of gravity.

Key concepts

Understanding the Derivative

The concept of a derivative is fundamental in calculus. It measures how a function changes as its input changes. In simpler terms, it's like checking how fast something is speeding up or slowing down at any given moment. For a position function, the derivative tells us the object's velocity.
The derivative is often represented by symbols like \(s'(t)\) or \(\frac{ds}{dt}\). It's the magic wand that changes position functions into velocity equations. Think of it as the speedometer of the mathematical world, showing us the exact speed (or velocity) at any point in time.
To find the derivative of a particular function like \(s(t) = -16t^2 + 500\), we use rules such as the power rule: \(\text{If } \ f(t) = t^n, \text{ then } \ f'(t) = n \cdot t^{n-1} \). This helps us systematically derive the expression for velocity.

Position Function and its Role

The position function describes where an object is at a particular time. It's a snapshot of the object's location as it varies over time, especially when affected by forces like gravity.
In the exercise, the position function is \(s(t) = -16t^2 + 500\), representing the height of an object falling from 500 feet.

• The \-16\ component reflects the pull of gravity affecting the object. It's important in determining how the object's speed changes over time.
• The \+500\ component represents the initial height from which the object was dropped.

This function provides a foundation for finding the object's velocity by taking its derivative, as the derivative converts the position information into velocity.

The Rate of Change Explained

Rate of change is all about understanding how one quantity changes in relation to another. It's like watching the hour hand on a clock move gradually over time. In mathematics, it's a crucial concept and is often derived using derivatives.
When applied to the position function, the rate of change tells us how the height of an object alters as time passes. It's the derivative stepping in again. In simple cases, it can be viewed as the slope of the line or curve representing the function. For our position function, the derivative \(s'(t)\) or \(-32t\)\ is the rate of change of the object's height.
So, the rate of change gives us a better grasp of how fast the velocity changes, indicating its real-world implications, like the speed with which a paint can falls.

Gravity's Impact on Motion

Gravity is a natural force pulling objects toward the center of the Earth. It's a constant force that affects motion and is essential to consider in physics problems.

• Gravity typically accelerates objects at a rate of approximately \(9.8 \, \text{m/s}^2\) or \(32 \, \text{ft/s}^2\) near the Earth's surface.
• In the position function \(s(t) = -16t^2 + 500\), the term \-16\ is related to this gravitational pull. It directly impacts how fast an object accelerates downwards.

By incorporating gravity into the position function, we accurately depict real-world scenarios such as a paint can dropping from height. Understanding gravity's role helps clarify why and how the velocity increases as an object falls, making it crucial for solving these kinds of problems.