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Chapter 2: Problem 5

If an average-size man with a parachute jumps from an airplane, he will fall \(12.5\left(0.2^{t}-1\right)+20 t\) feet in \(t\) seconds. How long will it take him to fall 140 feet?

Short Answer

Expert verified
It takes approximately 7 seconds for the parachute jumper to fall 140 feet.

Step by step solution

01

Understand the Problem

We need to find the time \( t \) it takes for a parachutist to fall 140 feet, given the formula for the distance fallen: \( d(t) = 12.5(0.2^{t} - 1) + 20t \). We are setting this formula equal to 140 and solving for \( t \).
02

Set Up the Equation

Set the given formula for distance equal to 140: \[12.5(0.2^{t} - 1) + 20t = 140\] This equation describes the relationship between time \( t \) and the distance fallen. Our task is to isolate \( t \).
03

Simplify the Equation

Start simplifying the equation. First, distribute the 12.5:\[12.5 \times 0.2^{t} - 12.5 + 20t = 140\]Add 12.5 to both sides to simplify:\[12.5 \times 0.2^{t} + 20t = 152.5\]
04

Trial and Error Method

Since the equation is transcendental and doesn't have an explicit solution method, we will use trial and error to find \( t \). Plug in values of \( t \) to see when the left side becomes close to 152.5. Try \( t = 4 \):\[12.5 \times 0.2^{4} + 20 \times 4 \]Calculate: \[12.5 \times 0.0016 + 80 = 12.5 \times 0.0016 + 80 = 0.02 + 80 = 80.02\]Since 80.02 is less than 152.5, try a higher value for \( t \).
05

Refine Estimate

Next, try \( t = 5 \):\[12.5 \times 0.2^{5} + 20 \times 5 \]Calculate: \[12.5 \times 0.00032 + 100 = 12.5 \times 0.00032 + 100 = 0.004 + 100 = 100.004\]100.004 is still less than 152.5. Try a larger value.
06

Find the Correct Time

Continue refining the guess. Try \( t = 6 \):\[12.5 \times 0.2^{6} + 20 \times 6\]Calculate:\[12.5 \times 0.000064 + 120 = 12.5 \times 0.000064 + 120 = 0.0008 + 120 = 120.0008\]Still too low. For \( t = 7 \):\[12.5 \times 0.2^{7} + 20 \times 7\]Calculating:\[12.5 \times 0.0000128 + 140 = 12.5 \times 0.0000128 + 140 = 0.00016 + 140 = 142.00016\]This is close to 140. After adjusting, \( t \approx 7 \, \text{seconds} \).
07

Verification

Verify by checking the calculations near \( t = 7 \). If \( t = 7 \) is too high, adjust downwards in small steps. The hint at \( t = 6.8) \, ext{seconds} \) balances closer to 140 feet.

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Key Concepts

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

Transcendental Equation
A transcendental equation involves a variable contained within a transcendental function. Common transcendental functions include exponential, logarithmic, and trigonometric functions. In these kinds of equations, the variable is not simply isolated on one side but is embedded within these functions. For example, in our parachutist exercise, the equation is:\[ 12.5(0.2^{t} - 1) + 20t = 140 \]Here, the variable \( t \) is present in the exponent. Additionally, unlike algebraic equations, transcendental equations do not always have simple or exact solutions. They often require numerical methods or approximations to solve. The nature of these equations makes them interesting but challenging, as straightforward algebraic techniques typically do not apply.
Trial and Error Method
The trial and error method is a useful strategy when dealing with equations that can't be solved using standard algebraic techniques. This approach involves testing different values to see which one satisfies the equation. In our exercise, we used trial and error because the presence of the exponential term \( 0.2^t \) in the equation made it difficult to isolate \( t \).This method can be summarized in these steps:
  • Choose an initial value for the variable.
  • Substitute this value into the equation.
  • Evaluate whether the equation is satisfied or needs adjustment.
  • Iterate with different values, refining progressively closer to the target solution.
Though not the most efficient method, especially for complex problems, it is effective for getting a close estimate quickly. Moreover, calculators or computer software can assist in speeding up this process.
Parachutist Distance Formula
The parachutist distance formula gives us a way to determine how far a parachutist will fall within a certain amount of time. The formula is:\[ d(t) = 12.5(0.2^{t} - 1) + 20t \]This equation combines two elements:
  • An exponential component \( 12.5(0.2^t) \) which models the initial resistance due to the parachute opening.
  • A linear component \( 20t \) representing the consistent rate of fall after the parachute begins to steady the descent.
This formula shows us the interplay of rapid initial deceleration due to the parachute's full deployment followed by a more stable, linear descent. Understanding this can be crucial not just for accurately predicting the fall time but also for analyzing how different factors and adjustments in a parachute system might affect descent rates.

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Most popular questions from this chapter

The manager of an employee health plan for a firm has studied the balance \(B\), in millions of dollars, in the plan account as a function of \(t\), the number of years since the plan was instituted. He has determined that the account balance is given by the formula \(B=60+7 t-50 e^{0.1 t}\). a. Make a graph of \(B\) versus \(t\) over the first 7 years of the plan. b. At what time is the account balance at its maximum? c. What is the smallest value of the account balance over the first 7 years of the plan?

The yearly profit \(P\) for a widget producer is a function of the number \(n\) of widgets sold. The formula is $$ P=-180+100 n-4 n^{2} . $$ Here \(P\) is measured in thousands of dollars, \(n\) is measured in thousands of widgets, and the formula is valid up to a level of 20 thousand widgets sold. a. Make a graph of \(P\) versus \(n\). b. Calculate \(P(0)\) and explain in practical terms what your answer means. c. What profit will the producer make if 15 thousand widgets are sold? d. The break-even point is the sales level at which the profit is 0 . Approximate the break-even point for this widget producer. e. What is the largest profit possible?

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One interesting problem in the study of dinosaurs is to determine from their tracks how fast they ran. The scientist R. McNeill Alexander developed a formula giving the velocity of any running animal in terms of its stride length and the height of its hip above the ground. 7 The stride length of a dinosaur can be measured from successive prints of the same foot, and the hip height (roughly the leg length) can be estimated on the basis of the size of a footprint, so Alexander's formula gives a way of estimating from dinosaur tracks how fast the dinosaur was running. See Figure \(2.57 .\) If the velocity \(v\) is measured in meters per second, and the stride length \(s\) and hip height \(h\) are measured in meters, then Alexander's formula is $$ v=0.78 s^{1.67} h^{-1.17} . $$ (For comparison, a length of 1 meter is \(39.37\) inches, and a velocity of 1 meter per second is about \(2.2\) miles per hour.) a. First we study animals with varying stride lengths but all with a hip height of 2 meters (so \(h=2\) ). i. Find a formula for the velocity \(v\) as a function of the stride length \(s\). ii. Make a graph of \(v\) versus \(s\). Include stride lengths from 2 to 10 meters. iii. What happens to the velocity as the stride length increases? Explain your answer in practical terms. iv. Some dinosaur tracks show a stride length of 3 meters, and a scientist estimates that the hip height of the dinosaur was 2 meters. How fast was the dinosaur running? b. Now we study animals with varying hip heights but all with a stride length of 3 meters (so \(s=3\) ). i. Find a formula for the velocity \(v\) as a function of the hip height \(h\). ii. Make a graph of \(v\) versus \(h\). Include hip heights from \(0.5\) to 3 meters. iii. What happens to the velocity as the hip height increases? Explain your answer in practical terms.
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