Problem 24 - Calculus Early Transcendentals

Question

A data collection probe is dropped from a stationary balloon and it falls with a velocity (in meters/second) given by \(v(t)=9.8 t,\) neglecting air resistance. After \(10 \mathrm{s},\) a chute deploys and the probe immediately slows to a constant speed of \(10 \mathrm{m} / \mathrm{s}\) which it maintains until it enters the ocean. a. Graph the velocity function. b. How far does the probe fall in the first 30 s after it is released? c. If the probe was released from an altitude of \(3 \mathrm{km},\) when does it enter the ocean?

Step-by-Step Solution

Answer

Answer: The probe falls 690 meters during the first 30 seconds and reaches the ocean after 261 seconds.

Step 1: Graphing the velocity function

To graph the velocity function, first create a graph of \(v(t)\) using the given equation \(v(t) = 9.8t\). Since the function is linear, plot the points \((0,0)\), which represents the initial condition, and \((10,98)\), which represents the point when the parachute is deployed. After 10s, the constant velocity is \(10\,\mathrm{m/s}\). The graph should show a straight line with a slope of \(9.8\) and the flat line at \(10\,\mathrm{m/s}\) after 10 seconds.

Step 2: Find the vertical distance after 30 s of free fall

To find the vertical distance covered during the first 30s, first find the distance covered during free fall, which lasts for 10 seconds. To do this, use the displacement formula \(s(t)=\int v(t) dt\) for the given velocity function \(v(t)=9.8t\) over the interval \([0,10]\). This will give you the distance fallen during free fall. \(s(t)=\int_{0}^{10} 9.8t\, dt = 9.8 \int_{0}^{10} t\, dt = 9.8 \left[\frac{1}{2}t^2\right]_0^{10}=9.8\left[\frac{1}{2} (10^2)\right]=490\) After the free fall, the velocity becomes constant at \(10\,\mathrm{m/s}\). The distance during the additional 20 seconds can be found using \(s(t)= vt\) as the velocity is constant. \(s(t)=10\cdot 20=200\) Now, sum both distances: the distance during free fall and the distance after the parachute deploys: \(Total\, distance= 490+200=690\,\mathrm{m}\) The probe falls 690 meters in the first 30 seconds after it is released.

Step 3: Calculate when the probe reaches the ocean

The probe is released from an altitude of 3 km, which is 3000 meters. We already know the distance traveled during the first 10 seconds of free fall and the velocity during the next phase. The remaining altitude can be found and used to find the remaining time to reach the ocean. \(Remaining\, altitude= 3000-490=2510\,\mathrm{m}\) Since the probe is moving at a constant speed of \(10\,\mathrm{m/s}\) after the chute deploys, we can find the time taken to cover the remaining distance as follows: \(Time = \frac{Remaining\, altitude}{Constant\, velocity}=\frac{2510}{10}=251\,\mathrm{s}\) Adding the 10 seconds of free fall and the 251 seconds during the constant speed phase, we get: \(Total\, time= 10+251=261\,\mathrm{s}\) The probe enters the ocean after 261 seconds.

Key Concepts

Velocity function

In kinematics, understanding the velocity function is crucial. It describes how fast an object moves and, importantly, how this speed changes over time. In our context, the velocity function given is:

• \( v(t) = 9.8t \) – this represents the velocity of the probe as a function of time, \( t \). This formula tells us that the probe accelerates at a constant rate of \( 9.8 \mathrm{m/s}^2 \), which is due to gravity.

This linear function is quite simple. At \( t = 0 \), the probe has no velocity since it was stationary at the start. As time increases, the velocity increases linearly. Imagine a graph with time on the horizontal axis and velocity on the vertical axis:

• At \( t = 10 \) seconds, the velocity reaches \( 98 \mathrm{m/s} \). This is when the parachute deploys.
• Beyond this point, velocity suddenly drops and becomes constant at \( 10 \mathrm{m/s} \), continuing at this speed until the probe hits the ocean.

This shift from increasing velocity to a constant one is pivotal in understanding how momentum and speed are managed by the parachute.

Displacement calculation

Displacement is key to understanding how far an object travels from its starting point over time. Calculating displacement with a velocity function involves integrating the velocity with respect to time. Initially, for our probe:

• During the first 10 seconds, where the velocity is given by \( v(t) = 9.8t \), we integrate to find the distance traveled: \[ s(t) = \int_{0}^{10} 9.8t\, dt = 490 \mathrm{m}.\]This tells us the probe falls 490 meters during free fall.
• After the parachute deploys, the velocity becomes constant at \( 10 \mathrm{m/s} \). So, for the following 20 seconds, displacement can be directly calculated as: \[ s(t) = 10 \times 20 = 200 \mathrm{m}. \]This adds up to a total displacement of 690 meters in the first 30 seconds.

By understanding displacement calculation, we see how integrals help describe motion, giving us the actual distance covered during varied phases of the probe's descent.

Graphing motion

Creating a graph of motion helps visualize how speed and distance evolve over time. For kinematics, the graph of the velocity function is particularly informative:

• The first segment of the graph, up to 10 seconds, forms a straight line. This slope of \( 9.8 \) reflects the consistent acceleration due to gravity, showing a continuous increase in speed.
• At 10 seconds, there's an abrupt change. The graph alters from a line to a horizontal segment, which represents motion at a constant velocity of \( 10 \mathrm{m/s} \).

Graphing motion involves plotting these elements:

• The initial segment, starting at the origin and reaching the point \((10, 98)\), captures the acceleration phase.
• The platform then levels as the parachute slows the probe to a consistent speed.

This graphical representation not only aids in understanding the physical journey of the probe but also highlights the importance of slopes and curves in depicting dynamic changes in motion.