Problem 13 - The Calculus with Analytic Geometry

Question

If an object falls from rest, its equation of motion is \(s=-16 t^{2}\), where \(t\) is the number of seconds in the time that has elapsed since the object left the starting point, \(s\) is the number of feet in the distance of the object from the starting point at \(t \mathrm{sec}\), and the positive direction is upward. If a stone is dropped from a building \(256 \mathrm{ft}\) high, find: (a) the instantaneous velocity of the stone 1 sec after it is dropped; (b) the instantaneous velocity of the stone 2 sec after it is dropped; (c) how long it takes the stone to reach the ground; (d) the instantaneous velocity of the stone when it reaches the ground.

Step-by-Step Solution

Answer

(a) -32 ft/sec, (b) -64 ft/sec, (c) 4 sec, (d) -128 ft/sec

Step 1: Understand the equation of motion

The equation of motion for a falling object is given by \[ s = -16t^2 \]

Step 2: Differentiate the equation to find velocity

The velocity \(v\) is the derivative of the displacement \( s \) with respect to time \( t \). Thus,\[ v(t) = \frac{ds}{dt} = \frac{d}{dt} (-16t^2) = -32t \]

Step 3: Calculate the instantaneous velocity at t = 1 sec

Substitute \( t = 1 \) into the velocity equation: \[ v(1) = -32(1) = -32 \text{ ft/sec} \]

Step 4: Calculate the instantaneous velocity at t = 2 sec

Substitute \( t = 2 \) into the velocity equation: \[ v(2) = -32(2) = -64 \text{ ft/sec} \]

Step 5: Set up the equation to find the time to reach the ground

The stone will hit the ground when \( s = -256 \). Set the equation equal to -256:\[ -16t^2 = -256 \]Solve for \( t \) by isolating \( t^2 \):\[ t^2 = 16 \]Then,\[ t = 4 \text{ sec} \]

Step 6: Calculate the instantaneous velocity when the stone hits the ground

Substitute \( t = 4 \) into the velocity equation: \[ v(4) = -32(4) = -128 \text{ ft/sec} \]

Key Concepts

derivatives in physics

Derivatives play an essential role in physics, particularly in understanding motion. One common application is to describe how an object's position changes over time and its velocity. When we differentiate the position function with respect to time, we get the velocity function.

In the example of a falling stone, our position equation is given by \(s = -16t^2\). To find the velocity, we take its derivative, which gives us \(v(t) = -32t\).

Derivatives help us understand instantaneous rates of change. In this case, the rate at which the stone's distance from the starting point changes with time.

equations of motion

Equations of motion describe the behavior of an object in terms of its displacement, velocity, and acceleration over time. These equations are crucial for solving problems involving moving objects.

Our example uses the equation \(s = -16t^2\), which represents the distance a stone falls over time. This equation assumes downward motion, hence the negative sign, with displacement measured in feet and time in seconds.

By differentiating this equation, we gain insights into how the stone's velocity changes over the elapsed time.

instantaneous velocity

Instantaneous velocity is the velocity of an object at a specific moment in time. It is the derivative of the position function.

For our falling stone, the velocity function derived from \(s = -16t^2\) is \(v(t) = -32t\). This tells us how fast and in which direction the stone is moving at any given time.

For example:

• At \(t = 1\) second, the instantaneous velocity is \(-32\) ft/sec
• At \(t = 2\) seconds, the instantaneous velocity is \(-64\) ft/sec
• When the stone hits the ground at \(t = 4\) seconds, its instantaneous velocity is \(-128\) ft/sec

problem-solving in calculus

Problem-solving in calculus often involves multiple steps and understanding the underlying concepts. Here’s a structured approach:

• Identify and understand the given equation or function, like \(s = -16t^2\)
• Differentiate the function to find related quantities, such as velocity, \(v(t) = \frac{ds}{dt} = -32t\)
• Substitute specific values to find exact rates, like \(v(1) = -32\) ft/sec for the instantaneous velocity at \(t = 1\) sec
• Set up equations for conditions given in the problem (e.g., finding when the stone reaches the ground by solving \(-16t^2 = -256\) leading to \(t = 4\) sec)

By following these steps, you can systematically tackle complex calculus problems related to motion and other phenomena.