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Chapter 6: Problem 4

Getting velocity from a formula: When a man jumps from airplane with an opening parachute, the distance \(S=S(t)\), in feet, that he falls in \(t \mathrm{sec}-\) onds is given by $$ S=20\left(t+\frac{e^{-1.6 t}-1}{1.6}\right) $$ a. Use your calculator to make a graph of \(S\) versus \(t\) for the first 5 seconds of the fall. b. Sketch a graph of velocity for the first 5 seconds of the fall.

Short Answer

Expert verified
Graph the function \( S(t) = 20\left(t+\frac{e^{-1.6 t}-1}{1.6}\right) \), then graph the velocity \( v(t) = 20(1 - e^{-1.6t}) \). Observe the increasing distance and velocity stabilizing over time.

Step by step solution

01

Understand the distance function

The given function for the distance \( S \) as a function of time \( t \) is \( S(t) = 20\left(t+\frac{e^{-1.6 t}-1}{1.6}\right) \). Our task is to graph this function and also determine its derivative to understand velocity.
02

Graph the distance function

Using a graphing calculator or a suitable graphing software, input the function \( S(t) = 20\left(t+\frac{e^{-1.6 t}-1}{1.6}\right) \). Plot this function for \( t \) ranging from 0 to 5 seconds.
03

Find the velocity function

Velocity \( v(t) \) is the derivative of the distance function \( S(t) \). Use the derivative rules to calculate \( \frac{dS}{dt} \). The derivative of \( S(t) \) is:\[\frac{dS}{dt} = 20\left(1 - e^{-1.6t}\right)\]
04

Graph the velocity function

Using a graphing calculator or software, plot the velocity function \( v(t) = 20\left(1 - e^{-1.6t}\right) \) over the interval from \( t = 0 \) to \( t = 5 \) seconds.
05

Interpret the graphs

The graph of \( S(t) \) will show an increasing curve as the man falls over time, reflecting increasing distance. The graph of \( v(t) \) will show the velocity increasing from 0 towards a stable value, indicating acceleration from rest until resistance from the parachute influences the speed.

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Key Concepts

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

Velocity
Velocity is a fundamental concept in calculus and physics, representing the rate at which an object's position changes with time. It is more than just speed; it also includes the direction of movement. In our exercise, the velocity is derived from the distance function, which tells us how far the man has fallen as a function of time.
  • Velocity is calculated as the derivative of the distance function, showing how quickly distance changes over a small time interval.
  • In our scenario, as the parachutist falls, his velocity starts at zero and increases as time goes on due to gravity, until the parachute begins to regulate the fall.

When graphing velocity, observe how it starts at zero and approaches a constant value, illustrating how the parachute gradually slows down further acceleration, flowing towards a steady descent.
Derivative
The derivative is a powerful mathematical tool used to find the rate of change of a quantity. It is the backbone of calculus and essential for understanding dynamics in various systems, such as the one given in the exercise.
  • In simple terms, the derivative tells us how a function's output value (y-value) changes as the input value (x-value) changes.
  • To find the derivative of the distance function, we apply the rules of differentiation to get the velocity function.
  • The derivative of our distance function is \( \frac{dS}{dt} = 20(1 - e^{-1.6t}) \), representing the velocity of the parachutist.

Understanding derivatives helps us make sense of not just how fast something moves, but how the speed itself changes over time, which is crucial for modeling real-world situations like parachuting.
Graphing Functions
Graphing functions allows us to visually analyze mathematical models, providing insights into how different variables interact. For this problem, we graph both the distance and velocity functions.
  • Graphing \( S(t) = 20(t + \frac{e^{-1.6 t}-1}{1.6}) \) gives us a view of how distance changes over the first 5 seconds of the fall.
  • When we graph the velocity \( v(t) = 20(1 - e^{-1.6t}) \), it shows us how speed changes, starting from zero and eventually reaching a plateau as the parachute opens.

By studying these graphs, we can intuitively understand the relationship between time, distance, and velocity, making complex calculus concepts accessible and relevant.
Mathematical Modeling
Mathematical modeling involves creating a mathematical representation of a real-world scenario in order to analyze and understand it better. This exercise exemplifies mathematical modeling perfectly.
  • The distance function provided models the parachutist's fall, accounting for gravitational pull and the parachute's resistance.
  • The velocity function derived from the distance function models how the speed of the fall changes with time.

This model allows us to predict future behavior, understand the effect of different forces (like gravity and parachute drag), and optimize situations, such as ensuring a safe descent. By using calculus and graphing, mathematical models transform abstract numbers and formulas into useful, real-world insights that guide our understanding of natural phenomena.

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Most popular questions from this chapter

Experimental determination of the drag coefficient: When retardation due to air resistance is proportional to downward velocity \(V\), in feet per second, falling objects obey the equation of change $$ \frac{d V}{d t}=32-r V $$ where \(r\) is known as the drag coefficient. One way to measure the drag coefficient is to measure and record terminal velocity. a. We know that an average-size man has a terminal velocity of 176 feet per second. Use this to show that the value of the drag coefficient is \(r=0.1818\) per second. (Hint: To say that the terminal velocity is 176 feet per second means that when the velocity \(V\) is 176 , velocity will not change. That is, \(\frac{d V}{d t}=0\). Put these bits of information into the equation of change and solve for \(r\).) b. An ordinary coffee filter has a terminal velocity of about 4 feet per second. What is the drag coefficient for a coffee filter?

A pond: Water is running out of a pond through a drainpipe. The amount of water, in gallons, in the pond \(t\) minutes after the water began draining is given by a function \(G=G(t)\). a. Explain the meaning in practical terms of \(\frac{d G}{d t}\). b. While water is running out of the pond, do you expect \(\frac{d G}{d r}\) to be positive or negative? c. When \(t=30\), water is running out of the drainpipe at a rate of 8000 gallons per minute. What is the value of \(\frac{d G}{d t}\) ? d. When \(t=30\), there are \(2,000,000\) gallons of water in the pond. Using the information from part c, estimate the value of \(G(35)\).

The rock with a formula: If from ground level we toss a rock upward with a velocity of 30 feet per second, we can use elementary physics to show that the height in feet of the rock above the ground \(t\) seconds after the toss is given by \(S=30 t-16 t^{2}\). a. Use your calculator to plot the graph of \(S\) versus \(t\). b. How high does the rock go? c. When does it strike the ground? d. Sketch the graph of the velocity of the rock versus time.

The acceleration due to gravity: From the time of Galileo, physicists have known that near the surface of the Earth, gravity imparts a constant acceleration of 32 feet per second per second. Explain how this shows that if air resistance is ignored, velocity for a falling object is a linear function of time.

Looking up: The constant \(g=32\) feet per second per second is the downward acceleration due to gravity near the surface of the Earth. If we stand on the surface of the Earth and locate objects using their distance up from the ground, then the positive direction is up, so down is the negative direction. With this perspective, the equation of change in velocity for a freely falling object would be expressed as $$ \frac{d V}{d t}=-g $$ (We measure upward velocity \(V\) in feet per second and time \(t\) in seconds.) Consider a rock tossed up- ward from the surface of the Earth with an initial velocity of 40 feet per second upward. a. Use a formula to express the velocity \(V=V(t)\) as a linear function. (Hint: You get the slope of \(V\) from the equation of change. The vertical intercept is the initial value.) b. How many seconds after the toss does the rock reach the peak of its flight? (Hint: What is the velocity of the rock when it reaches its peak?) c. How many seconds after the toss does the rock strike the ground? (Hint: How does the time it takes for the rock to rise to its peak compare with the time it takes for it to fall back to the ground?)
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