Problem 41 - Applied Calculus

Question

If you jump out of an airplane and your parachute fails to open, your downward velocity (in meters per second) \(t\) seconds after the jump is approximated by $$ v(t)=49\left(1-(0.8187)^{t}\right) $$ (a) Write an expression for the distance you fall in \(T\) seconds. (b) If you jump from 5000 meters above the ground, estimate, using trial and error, how many seconds you fall before hitting the ground.

Step-by-Step Solution

Answer

(a) The distance function is \( s(t) = 49t - \frac{49}{\ln(1/0.8187)} (0.8187)^{t} \). (b) Estimate \( T \approx 92 \) seconds to fall from 5000 meters.

Step 1: Understand the Velocity Function

The given velocity function of a falling object is \( v(t) = 49 \left(1 - (0.8187)^{t}\right) \). This function represents the downward velocity (in meters per second) of a parachute jumper whose parachute has failed, at any time \( t \) seconds.

Step 2: Find Distance as a Function of Time

The distance fallen as a function of time can be found by integrating the velocity function with respect to time. Therefore, the distance function \( s(t) \) is given by \( s(t) = \int v(t) \, dt = \int 49 \left(1 - (0.8187)^{t}\right) \, dt \).

Step 3: Perform the Integration

Integrate the velocity function: Given \( v(t) = 49 \left(1 - (0.8187)^{t}\right) \), we can integrate it:\(\int v(t) \, dt = \int 49 \, dt - \int 49 \, (0.8187)^{t} \, dt.\)The first part integrates to \( 49t \), and the second part requires a substitution for the exponential, resulting in: \( s(t) = 49t - \frac{49}{\ln(1/0.8187)} (0.8187)^{t} + C \).For simplicity, assume \( C = 0 \) since our initial point \( s(0) = 0 \).

Step 4: Use the Distance Function to Find Time to Hit the Ground

Set \( s(t) = 5000 \) meters to determine the time when the jumper hits the ground. We solve:\( 49T - \frac{49}{-\ln(0.8187)} (0.8187)^{T} = 5000 \) Use trial and error or numerical methods to find the value of \( T \) that satisfies the equation. You can use a calculator or software to approximate this numerically.

Key Concepts

Velocity Function

In the context of a skydiver's free fall, the velocity function helps us understand how fast the skydiver is moving at any given time after jumping out of the plane. The specific function provided in the exercise is:
\[ v(t) = 49 \left(1 - (0.8187)^t\right) \]This function gives the velocity in meters per second, where \( t \) is the time in seconds. Here are a few key aspects to take away:

• **Understanding the Expression**: The term \( 49 \) represents the maximum velocity the jumper can achieve as terminal velocity approaches. The expression \((1 - (0.8187)^t)\) indicates how quickly the velocity reaches this maximum speed.
• **Exponential Decay**: The factor \( (0.8187)^t \) shows an exponential approach towards zero. As time increases, the influence of \( (0.8187)^t \) becomes negligible, causing the velocity to near 49 m/s.
• **Implications in Real Life**: In a real fall without a parachute, the velocity function reflects how air resistance affects speed, though this particular model assumes both parachute failure and air resistance parameters already considered.

Understanding this function is crucial as it forms the basis for determining how far the skydiver has fallen over time.

Distance Function

To find how far the skydiver falls, we translate the velocity function into a distance function using integration. This gives us the total distance fallen as time progresses:
The process of integration transforms the rate of change—velocity—into a position over time—distance. The distance function \( s(t) \) can be expressed as:
\[ s(t) = \int v(t) \, dt = \int 49 \left(1 - (0.8187)^t\right) \, dt \]After performing the integration step by step, we find:

• **Basic Integration**: The integral of a constant \(49\) with respect to time \(t\) is \(49t\).
• **Integrating the Exponential Term**: To solve \(\int (0.8187)^t \, dt\), substitution techniques yield \(\frac{49}{\ln(1/0.8187)} (0.8187)^t\).
• **Resulting Function**: Combining these gives the complete distance function:\[ s(t) = 49t - \frac{49}{-\ln(0.8187)} (0.8187)^t + C \]In this scenario, \(C\) is initially considered zero because at \(t = 0\), the distance falls \(s(0) = 0\).

This distance function is essential for calculating the fall's total distance over a given period.

Integration

Integration is a fundamental process in calculus that allows us to find the area under a curve. It is the reverse operation of differentiation and plays a pivotal role in converting the velocity function to a distance function.

• **Purpose**: In the context of a falling object, integration helps determine the total distance fallen by summing up all small increments of displacement over time.
• **Mechanics of Integration**: When integrating a function like \( v(t) = 49(1 - (0.8187)^t) \), we separate it into simpler parts to handle individually: \[\int 49 \, dt - \int 49 \times (0.8187)^t \, dt \]
• **Exponential Integrals**: The integral of an exponential function such as \((0.8187)^t\) requires a specific technique involving natural logarithms. This is why you see terms like \(\frac{49}{\ln(1/0.8187)}\) appear in the final expression.
• **Constant of Integration**: Every integration includes a constant \(C\), accounting for initial conditions. For simplicity, we often let \(C=0\) in physics when the context is clear, like starting from rest or the ground level.

By understanding integration, you can see how changes in speed integrate into total distance, bridging the gap between dynamics and kinematics.