Problem 14 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A squirrel falls from a tree from a height of 10 meters above the ground. At time \(t\) seconds after it slips from the tree, the squirrel is a distance \(s(t)=10-4.9 t^{2}\) meters above the ground. How fast is the squirrel falling when it hits the ground?

Step-by-Step Solution

Answer

The squirrel is falling at approximately 14.0 meters per second downward when it hits the ground.

Step 1: Understand the problem

We need to find the speed of the squirrel when it hits the ground. This involves finding the derivative of the position function which gives the velocity of the squirrel, and evaluating it at the time it hits the ground.

Step 2: Determine when the squirrel hits the ground

The squirrel hits the ground when its height above the ground is 0 meters. Set the equation for distance equal to 0: \[ 10 - 4.9t^2 = 0 \]Solve for \( t \).

Step 3: Solve for time when squirrel hits the ground

Rearrange the equation to find \( t \): \[ 4.9t^2 = 10 \]\[ t^2 = \frac{10}{4.9} \]\[ t = \sqrt{\frac{10}{4.9}} \]

Step 4: Calculate time

Calculate the value of \( t \):\[ t \approx \sqrt{2.04} \approx 1.428 \] seconds. This is the time when the squirrel hits the ground.

Step 5: Find the velocity function

The velocity of the squirrel is given by the derivative of the position function. Differentiate \( s(t) = 10 - 4.9t^2 \) with respect to \( t \): \[ v(t) = \frac{ds}{dt} = -9.8t \]

Step 6: Calculate the velocity when squirrel hits the ground

Substitute \( t = 1.428 \) seconds into the velocity function to find the speed:\[ v(1.428) = -9.8 \times 1.428 \approx -14.0 \] meters per second. The negative sign indicates the direction is downward.

Key Concepts

Position Function

The position function is a mathematical expression that describes the location of an object at a given time. In this exercise, the position of the falling squirrel is represented by the function \[ s(t) = 10 - 4.9t^2 \] where \(s(t)\) is the height above the ground in meters, and \(t\) is the time in seconds. This specific quadratic equation is characteristic of objects in free-fall under gravity, starting from a certain height.

• The initial position, 10 meters, indicates the starting height from which the squirrel falls.
• The term \( -4.9t^2 \) represents the effect of gravity, causing the squirrel to accelerate downwards.

Understanding the position function is crucial because it not only tells us where the squirrel is at any given time but also serves as the foundation for finding other related quantities like velocity.

Velocity

Velocity is the rate at which an object's position changes with respect to time. In this problem, it represents how quickly the squirrel's position is changing as it falls. Once we know the position function, the velocity can be found by differentiating the position function with respect to time. For the given function:\[ v(t) = \frac{ds}{dt} \]In our case, differentiating \[ s(t) = 10 - 4.9t^2 \] with respect to \(t\), gives us: \[ v(t) = -9.8t \]

• This negative sign in \(v(t)\) indicates a direction opposite to the reference positive direction; here, downwards.
• When the velocity is calculated exactly when the squirrel reaches the ground, we see how fast the squirrel is moving at that critical moment.

Derivative

A derivative in calculus is a critical tool used to determine how a function is changing at any given point. Specifically, it measures the rate of change of a function with respect to one of its variables. In this exercise, finding the derivative of the position function, \[ s(t) = 10 - 4.9t^2 \],allows us to determine the velocity of the falling squirrel. Here, the derivative expresses how the height of the squirrel changes over time:\[ \frac{ds}{dt} = -9.8t \]

• The derivative, \(-9.8t\), quantifies how quickly the squirrel is descending over time.
• It effectively converts positional information (in meters) to speed information (in meters per second), crucial for understanding motion.

By understanding and applying derivatives, we can predict future positions and velocities of moving objects like the squirrel.

Distance-Time Relationship

The distance-time relationship is a key concept in physics and calculus, explaining how an object's position or distance from a reference point changes over time. In scenarios like the falling squirrel, this relationship helps to quantify movement and motion.

• The function \(s(t) = 10 - 4.9t^2\) describes how the distance above the ground decreases with time as the squirrel falls.
• By setting \(s(t)\) to zero, we determine the exact time when the squirrel reaches the ground. Solving \(10 - 4.9t^2 = 0\) gives us \(t \approx 1.428\) seconds.

Through the derivative of this function, we connect distance with speed, gaining insights into how fast the squirrel falls as it approaches the ground. This understanding is essential in analyzing and predicting motion.