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Chapter 11: Problem 23

A ball is dropped from a boat so that it strikes the surface of a lake with a speed of \(16.5 \mathrm{ft} / \mathrm{s}\). While in the water the ball experiences an acceleration of \(a=10-0.8 v,\) where \(a\) and \(v\) are expressed in flus' and fts, respectively. Knowing the ball takes \(3 \mathrm{s}\) to reach the bottom of the lake, determine \((a)\) the depth of the lake, \((b)\) the speed of the ball when it hits the bottom of the lake.

Short Answer

Expert verified
The lake's depth is 34.6 ft, and the ball's speed at the bottom is 12.57 ft/s.

Step by step solution

01

Define the equations

We are given an acceleration function for the ball in the water: \( a = 10 - 0.8v \). This is a differential equation relating acceleration to velocity.
02

Set up the velocity equation

Since acceleration is the derivative of velocity with respect to time, we can rewrite the relationship as: \( \frac{dv}{dt} = 10 - 0.8v \).
03

Solve the differential equation

This is a separable differential equation. First, rearrange it to integrate:\[ \frac{dv}{10 - 0.8v} = dt \]Integrating both sides:\[ \int \frac{1}{10 - 0.8v} \, dv = \int dt \]This gives us:\[ -\frac{1}{0.8} \ln|10 - 0.8v| = t + C \]
04

Solve for velocity as a function of time

Exponentiate both sides and simplify:\[ 10 - 0.8v = Ce^{-0.8t} \]\[ v = \frac{10 - Ce^{-0.8t}}{0.8} \]
05

Apply initial conditions to find constants

When \( t = 0 \), the velocity \( v = 16.5 \). Substitute these values into the equation:\[ 16.5 = \frac{10 - C}{0.8} \]Solve for \( C \):\[ C = 10 - 0.8 \times 16.5 \]
06

Solve for C and substitute back

Calculating \( C \):\[ C = 10 - 13.2 = -3.2 \]So, the velocity function becomes:\[ v(t) = \frac{10 + 3.2e^{-0.8t}}{0.8} \]
07

Find speed at the bottom of the lake

At \( t = 3 \) s, find \( v \):\[ v(3) = \frac{10 + 3.2e^{-2.4}}{0.8} \]
08

Compute v(3) numerically

Evaluate:\[ v(3) \approx \frac{10 + 3.2e^{-2.4}}{0.8} \approx 12.57 \text{ ft/s} \]
09

Set up and solve the depth equation

Distance \( x \) is the integral of velocity:\[ x(t) = \int v(t) \, dt = \int \left(\frac{10 + 3.2e^{-0.8t}}{0.8}\right) \, dt \]
10

Find the depth of the lake

Perform the integration and evaluate from \( t=0 \) to \( t=3 \):\[ x(3) = \left(\frac{10}{0.8}t - \frac{3.2}{0.64}e^{-0.8t}\right)\bigg|_{0}^{3} \]Perform the calculations to find the depth.
11

Calculate x(3) numerically

Solve the integral:\[ x(3) = \left(12.5 \times 3 - \frac{3.2}{0.64} (e^{-2.4}-1)\right) \approx 34.6 \text{ ft} \]
12

Conclusion

The depth of the lake is approximately \( 34.6 \text{ ft} \), and the speed of the ball when it hits the bottom is approximately \( 12.57 \text{ ft/s} \).

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Key Concepts

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

Understanding Kinematics
Kinematics is a branch of mechanics that focuses on the motion of objects without considering the forces causing that motion. In this exercise involving a ball dropped into a lake, kinematics helps us understand various aspects of its motion. We start by examining what happens as the ball moves through the water.
  • Kinematics provides tools to describe motion using quantities such as displacement, velocity, and acceleration.
  • The kinematic equations allow us to connect these quantities, offering a way to analyze an object's movement comprehensively.
  • Understanding kinematics involves learning how to describe motion accurately using graphs and calculations.
In our problem, we utilize kinematics to assess how the ball's speed changes over time and how far it travels to reach the bottom of the lake. This forms the foundation for exploring related concepts like velocity and acceleration.
Exploring Acceleration
Acceleration is a key part of understanding motion, particularly when movement involves changes in speed. In our example, the ball experiences specific acceleration as it drops into the water, defined by the equation: \[ a = 10 - 0.8v \]This equation reveals a crucial relationship between acceleration and velocity.
  • Acceleration measures how quickly the velocity of an object changes over time.
  • Positive values imply speeding up, while negative values suggest slowing down.
  • Here, the acceleration is not constant; it's influenced by the ball's speed, making the situation more dynamic.
This relationship signifies that as the ball speeds up underwater, the acceleration decreases, eventually balancing until it reaches a steady velocity. This aspect of acceleration is vital in solving the differential equations for this problem.
Determining Velocity
Velocity tells us not only how fast an object moves but in which direction. Solving the differential equation given in the problem provides insight into how the velocity of the ball changes over time:
\[ v(t) = \frac{10 + 3.2e^{-0.8t}}{0.8} \]
  • The equation describes velocity as a function of time, accounting for initial conditions like the initial release speed.
  • Velocity is a vector quantity, including both magnitude and direction, though in this scenario, the direction is implicit as the ball moves downwards only.
  • Knowing the velocity at specific moments (e.g., after 3 seconds) helps us understand how incidents like hitting the lake's bottom occur.
In this case, calculating velocity after 3 seconds reveals that the ball reaches about 12.57 ft/s鈥攄emonstrating the impact of the changing acceleration underwater. This velocity insight aids in understanding not just the movement but also overall travel dynamics.
Calculating Depth
To determine how deep the ball travels, we compute the total distance it covers until it reaches the lake's bottom using an integration of the velocity function:
\[ x(t) = \int v(t) \, dt \]By performing this integration from time 0 to 3 seconds, we find the depth:
\[ x(3) = \left( \frac{10}{0.8}t - \frac{3.2}{0.64}e^{-0.8t} \right)\bigg|_{0}^{3} \]
  • Integration helps convert velocity over time into a tangible measure of distance.
  • It considers the initial and final points, accounting for the change from surface to deep water.
  • Depth calculation is pivotal in understanding the ball鈥檚 displacement.
With calculated results, we determine the lake's depth to be roughly 34.6 feet. This final depth evaluation offers closure on the challenge, uniting the trajectory data into clear physical properties.

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Most popular questions from this chapter

Based on observations, the speed of a jogger can be approximated by the relation \(v=7.5(1-0.04 x)^{0.3}\), where \(v\) and \(x\) are expressed in milh and miles, respectively. Knowing that \(x=0\) at \(t=0\), determine \((a)\) the distance the jogger has run when \(t=1 \mathrm{h},(b)\) the jogger's acceleration in fts at \(t=0,(c)\) the time required for the jogger to run \(6 \mathrm{mi}\).

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