Question

The distance an object falls (when released from rest, under the influence of Earth's gravity, and with no air resistance) is given by \(d(t)=16 t^{2},\) where \(d\) is measured in feet and \(t\) is measured in seconds. A rock climber sits on a ledge on a vertical wall and carefully observes the time it takes a small stone to fall from the ledge to the ground. a. Compute \(d^{\prime}(t) .\) What units are associated with the derivative and what does it measure? Interpret the derivative. b. If it takes 6 s for a stone to fall to the ground, how high is the ledge? How fast is the stone moving when it strikes the ground (in \(\mathrm{mi} / \mathrm{hr}\) )?

Short answer

Answer: The height of the ledge is 576 feet, and the stone strikes the ground at approximately 130.9 miles per hour.

Step-by-step solution

Compute \(d'(t)\)

To find the first derivative of \(d(t)\) with respect to time, we can apply the power rule for differentiation: \[ \frac{d}{dt}(t^n) = nt^{n-1} \] Here, \(n=2\). Therefore, \[ d'(t) = \frac{d}{dt}(16t^2) = 16 \cdot 2t^{2 - 1} = 32t. \]

Calculate units and interpret the derivative

The derivative represents the rate of change of the function with respect to time. In this case, \(d'(t)\) represents the velocity of the object. Since \(d(t)\) is measured in feet and time is measured in seconds, the units of \(d'(t)\) will be feet per second (ft/s). The physical interpretation of the derivative is that it gives the stone's velocity at any given time-\(t\).

Compute the height of the ledge

We are given that it takes 6 seconds for the stone to fall to the ground. To find the height of the ledge, we need to substitute this value of time into the distance function \(d(t)\). Thus, \[d(6) = 16(6)^2 = 16(36) = 576\, \text{ft} \] The height of the ledge is 576 feet.

Compute the stone's speed when it strikes the ground

To find the stone's speed when it hits the ground, we must substitute the given time 6 seconds into the velocity function, \(d'(t)\): \[ d'(6) = 32(6) = 192 \, \text{ft/s} \] Now, we need to convert this value from feet per second to miles per hour: \[\frac{192 \, \text{ft/s}}{1} \times \frac{1\, \text{mile}}{5280\, \text{ft}} \times\frac{3600\, \text{s}}{1\, \text{hr}} \approx 130.9 \, \text{mi/hr}\] The stone is moving at approximately 130.9 miles per hour when it strikes the ground.

Key concepts

Power Rule Differentiation

Understanding the power rule for differentiation is crucial in calculus, especially when dealing with physics problems involving motion. The power rule states that the derivative of a variable raised to a power is equal to the power multiplied by the variable raised to one less than that power. In the language of mathematical notation: \[\begin{equation}\frac{d}{dx}(x^n) = nx^{n-1}\text{,}\text{ where } n \text{ is a constant.}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }어 클램프가는\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\end{equation}\]For the rock climber's problem, when applying the power rule to the given quadratic function, we differentiate the term involving time squared (\(t^2\)) to find the velocity of the stone. Using this method simplifies complex calculus problems and allows for a straightforward interpretation of physical situations. In this case, the power rule shows us how the velocity of a falling object changes over time.

Velocity and Acceleration

In physics, understanding the concepts of velocity and acceleration is essential. Velocity is the rate at which an object changes its position over time. It is a vector quantity, meaning it has both magnitude and direction. Acceleration, on the other hand, is the rate of change of velocity over time.When we take the derivative of the distance function with respect to time, as in the problem at hand, we obtain the velocity function. In our example, the derivative was found to be \(d'(t) = 32t\), which means the velocity increases linearly with time (since there is no air resistance). This is expected for an object in free fall under the influence of gravity, which is a perfect example of uniformly accelerated motion.After determining the velocity of the stone at the moment it hits the ground, we can better understand the motion involved. The constant rate of acceleration due to gravity is implied in the function given, and the exercise demonstrates the direct proportionality between time and velocity in the case of constant acceleration.

Units Conversion

In science and engineering, units conversion is necessary for interpreting results in different systems of measurement. The process involves using conversion factors to switch from one unit to another. In our example, we convert the stone's velocity from feet per second to miles per hour.To do so, we use the conversion factors that 1 mile equals 5280 feet and 1 hour equals 3600 seconds. Such conversion is essential when the context requires specific units, such as miles per hour in vehicle speedometers. Knowing how to properly convert units can prevent misunderstandings and errors in various fields, not just physics. It's important to ensure that the conversion factors are used in the correct orientation to preserve the meaning of the original measurement.

Quadratic Functions

Quadratic functions are polynomial functions of the second degree, generally in the form \(f(x) = ax^2 + bx + c\). These functions represent parabolas when graphed, and they frequently appear in physics and mathematics. The distance-versus-time function in our problem, \(d(t) = 16t^2\), is a quadratic function.Quadratics are particularly relevant in physics problems that involve projectiles or objects in free fall (as in our rock climber’s example), where the squared term represents the object's acceleration due to gravity. Solving quadratic functions allows us to find valuable information like the maximum height of a projectile or the time it takes for an object to reach the ground. Understanding the properties and applications of quadratic functions is an essential tool for analyzing motion and predicting outcomes.