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Chapter 3: Problem 51

The height of an object \(t\) seconds after it is dropped from a height of 300 meters is \(s(t)=-4.9 t^{2}+300\) . (a) Find the average velocity of the object during the first 3 seconds. (b) Use the Mean Value Theorem to verify that at some time during the first 3 seconds of fall, the instantaneous velocity equals the average velocity. Find that time.

Short Answer

Expert verified
Part (a) gives us that the average velocity is -29.4 m/s. From part (b), we can verify that there is a point where the instantaneous velocity is equal to -29.4 m/s, specifically at time \(t = 3\) seconds.

Step by step solution

01

Calculate the Average Velocity

To calculate the average velocity of the object over the time interval from \(t=0\) to \(t=3\), use the formula \(\frac{s(b) - s(a)}{b - a}\). Here, a stands for the initial time and b for the final time. Hence, the average velocity will be \(\frac{s(3)-s(0)}{3-0}\).
02

Plug into the Equation

From step 1, we know that \(s(t)=-4.9 t^{2}+300\), hence \(s(3)=-4.9 \cdot 3^{2}+300\) and \(s(0)=-4.9 \cdot 0^{2}+300\). Calculate these values and substitute back into the equation \(\frac{s(3)-s(0)}{3-0}\).
03

Use the Mean Value Theorem

According to the Mean Value Theorem, if a function \(s(t)\) is continuous on the interval [a, b] and differentiable on the interval (a, b), there exists a number c in the interval (a, b) where the derivative of the function at c equals the average rate of change of the function over [a, b]. Expressing the Mean Value Theorem using the current situation gives us \(s'(c)=\frac{s(3)-s(0)}{3-0}\). It is known that \(s(t)=-4.9 t^{2}+300\), so \(s'(t)=-9.8 t\). We can solve for \(c\) by setting \(s'(c)\) equal to the average velocity we obtained in Step 2.
04

Solve for the Instantaneous Velocity

Solve for \(c\) in the equation \(s'(c)=\frac{s(3)-s(0)}{3-0}\) using the value for average velocity from Step 2. The resulting \(c\) value will provide the time where the instantaneous velocity matches the average velocity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Velocity Calculus
In calculus, the average velocity represents the rate of change in position over a specific time interval. It's calculated simply by taking the difference in position values at two distinct times and dividing by the time interval. For the object described in the exercise, with its position given by the equation \( s(t) = -4.9t^2 + 300 \), we find the average velocity over the first 3 seconds by subtracting the initial height from the height after 3 seconds and dividing by 3.
To visualize this, let's suppose you took a trip from your house to a park and back. If you know the distance travelled and the total time taken, you could quickly calculate the average speed of your whole trip using this same technique. However, average velocity doesn't tell us about variations in speed during the trip 鈥� it simply provides a 'big picture' overview.
Instantaneous Velocity Calculus
Instantaneous velocity, unlike average velocity, gives us the speed of an object at a specific moment in time. It's derived by evaluating the derivative of the position function at a point, which in the realm of calculus, is the slope of the tangent line at that point on the position-time graph. For our falling object, we find the instantaneous velocity by differentiating \( s(t) \) to get \( s'(t) = -9.8t \).

Finding the Moment When Instantaneous Equals Average Velocity

Using the Mean Value Theorem, we can establish that there must be a point within a time interval where the instantaneous velocity equals the average velocity. By solving \( s'(c) = \frac{s(3)-s(0)}{3-0} \), we determined this moment, which offers a snapshot of the object's speed鈥攎uch like capturing the precise speed of a car at an instant using a speedometer.
Differential Calculus
Differential calculus lies at the heart of understanding rate changes, and it's what powers the calculations of both average and instantaneous velocities. It involves finding derivatives, which are the rates of change of functions. The process of differentiation allows us to move from the general formula for the position of the object \( s(t) \) to its velocity function \( s'(t) \).

In real-world applications, differential calculus is used everywhere from physics to optimize functions in economics or calculating the rate at which a disease spreads in epidemiology. Its versatility and practicality make it a cornerstone for understanding dynamic systems in a precise, quantitative way.

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Most popular questions from this chapter

Circumference The measurement of the circumference of a circle is found to be 64 centimeters, with a possible error of 0.9 centimeter. (a) Approximate the percent error in computing the area of the circle. (b) Estimate the maximum allowable percent error in measuring the circumference if the error in computing the area cannot exceed \(3 \% .\)

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Numerical, Graphical, and Analytic Analysis An exercise room consists of a rectangle with a semicircle on each end. A 200 -meter running track runs around the outside of the room. (a) Draw a figure to represent the problem. Let \(x\) and \(y\) represent the length and width of the rectangle. (b) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) Use the table to guess the maximum area of the rectangular region. $$ \begin{array}{|c|c|c|}\hline \text { Length, } x & {\text { Width, } y} & {\text { Area, } x y} \\ \hline 10 & {\frac{2}{\pi}(100-10)} & {(10) \frac{2}{\pi}(100-10) \approx 573} \\ \hline 20 & {\frac{2}{\pi}(100-20)} & {(20) \frac{2}{\pi}(100-20) \approx 1019} \\ \hline\end{array} $$ (c) Write the area \(A\) as a function of \(x\) . (d) Use calculus to find the critical number of the function in part (c) and find the maximum value. (e) Use a graphing utility to graph the function in part (c) and verify the maximum area from the graph.
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