Problem 49 - Essential Calculus Early Transcendentals

Question

since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 \(\mathrm{m} / \mathrm{s}\) and its downward acceleration is $$a=\left\\{\begin{array}{ll}{9-0.9 t} & {\text { if } 0 \leqslant t \leqslant 10} \\ {0} & {\text { if } t>10}\end{array}\right.$$ If the raindrop is initially 500 \(\mathrm{m}\) above the ground, how long does it take to fall?

Step-by-Step Solution

Answer

The raindrop takes approximately 15.56 seconds to fall to the ground.

Step 1: Understand the Problem

The problem gives us an initial downward velocity and a piecewise function for acceleration. Our goal is to find out how long it takes the raindrop to reach the ground from an initial height of 500 meters.

Step 2: Integrate the Acceleration to Find the Velocity Function

The acceleration given is a piecewise function: \( a(t) = 9 - 0.9t \) for \( 0 \leq t \leq 10 \) and \( a(t) = 0 \) for \( t > 10 \). Integrate \( a(t) \) from \( 0 \) to \( t \) to find the velocity function for \( t \leq 10 \): \( v(t) = \int (9-0.9t) \, dt = 9t - 0.45t^2 + C \). Given that the initial velocity at \( t = 0 \) is 10 m/s, we can solve for \( C \) using \( v(0) = 10 \), leading to \( C = 10 \). Therefore, \( v(t) = 9t - 0.45t^2 + 10 \) for \( 0 \leq t \leq 10 \). After \( t > 10 \), \( v(t) \) remains constant because acceleration is zero.

Step 3: Integrate the Velocity to Find the Position Function

Since we have \( v(t) = 9t - 0.45t^2 + 10 \) for \( 0 \leq t \leq 10 \), integrate \( v(t) \) to find the position function \( s(t) \). This gives \( s(t) = \int (9t - 0.45t^2 + 10) \, dt = \frac{9}{2}t^2 - 0.15t^3 + 10t + C \). Given that \( s(0) = 500 \) meters (initial height), solve for \( C \): \( C = 500 \). Hence, \( s(t) = \frac{9}{2}t^2 - 0.15t^3 + 10t + 500 \). For \( t > 10 \), the position function is a linear extension of the final velocity.

Step 4: Calculate the Time for Ground Impact for \(0 \leq t \leq 10\)

Determine if the raindrop hits the ground within the first 10 seconds by solving \( s(t) = 0 \). Substitute in the position function: \( \frac{9}{2}t^2 - 0.15t^3 + 10t + 500 = 0 \). Solve this cubic equation for \( t \), which is complex and can be done numerically. If \( t \) is greater than 10, proceed to next step.

Step 5: Calculate the Continued Descent After \(t = 10\)

If the drop doesn't hit within 10 seconds, use \( t = 10 \) final position and velocity. Evaluate \( s(10) = \frac{9}{2}(10)^2 - 0.15(10)^3 + 10(10) + 500 = 250 \) meters. Now, the velocity is constant \( v = 45 \) m/s for \( t > 10 \). Find \( t' \) such that \( 45t' = 250 \) to reach the ground. Solve for \( t' \), giving \( t' = \frac{250}{45} \approx 5.56 \) seconds.

Step 6: Total Time for Impact

To find the total time for impact with the ground, sum the time intervals: \( t = 10 + 5.56 = 15.56 \) seconds. This accounts for initial part of motion with changing acceleration and later constant velocity descent.

Key Concepts

Piecewise Functions

In the context of calculus and kinematics, piecewise functions are used to represent scenarios where different conditions apply within certain intervals. A piecewise function is characterized by distinct expressions defining it over different portions of the domain.

In our raindrop problem, the acceleration function of the raindrop is given in a piecewise form:

• For the interval \(0 \leq t \leq 10\), the acceleration is described by the equation \(a(t) = 9 - 0.9t\).
• Once \(t\) exceeds \(10\) seconds, the acceleration drops to \(0\), meaning the raindrop moves at constant velocity.

By understanding how to interpret piecewise functions, we can analyze how different physical forces affect motion over time. Recognizing these changes allows for accurate mathematical modeling.

Integration Techniques

Integration is a fundamental concept in calculus, particularly useful in finding antiderivatives or the area under curves. It's essential in solving problems involving motion, as it helps determine velocity and position.

• To solve our problem, we begin by integrating the acceleration function, \(a(t) = 9 - 0.9t\), to obtain the velocity function, \(v(t)\), for \(0 \leq t \leq 10\).
• Applying basic integration gives us \(v(t) = \int (9 - 0.9t) \, dt = 9t - 0.45t^2 + C\).

The integration constant, \(C\), is determined using initial conditions, such as the initial velocity of the raindrop when \(t = 0\).

Integrating the velocity function then helps us find the position function \(s(t)\), crucial for determining how far the raindrop travels over time.

Kinematics Analysis

Kinematics is a branch of physics that describes motion without considering the forces that cause it. In this exercise, analyzing the raindrop's motion involves:

• Initial velocity: The raindrop starts with a velocity of 10 m/s downward.
• Acceleration: The raindrop experiences a changing acceleration due to air resistance increasing with surface area.
• Piecewise motion: It includes separate motion phases, with initially decreasing acceleration and then constant velocity.

By understanding these components, we piece together the raindrop's journey from its initial drop to its contact with the ground. This requires combining both equations and intuitive understanding of how changing conditions (acceleration to zero) affect the descent.

Differential Equations

Differential equations play a crucial role in calculating real-world phenomena, especially in motion analysis. They relate functions and their derivatives, forming the basis for finding unknown functions.

In our scenario, \(a(t)\) as a function involved solving a differential equation:

• The differential form for the velocity, \(v(t)\), is solved by integrating \( a(t) \).
• The second integration, \(\int v(t) \, dt\), gives the position \(s(t)\), another differential equation step.

Both procedures require initial conditions for precise solutions, like the starting velocity and position in the raindrop example. Understanding differential equations bridges initial data with overall behavior prediction.

Velocity and Acceleration

Velocity and acceleration are key concepts in kinematics, describing how an object's speed and direction of movement change.

• Velocity (\(v(t)\)) represents how fast and in which direction the raindrop moves. It's derived by integrating the acceleration function.
• Acceleration (\(a(t)\)) illustrates how velocity changes over time. Initially, it's non-zero and changes due to atmospheric resistance.

For our problem, understanding these variations helps outline the motion from free fall to reaching the ground with constant velocity. Through integration, we determine the velocity function's impact on the object's descent, building up a clear timeline of motion.