Problem 52 - Calculus an Applied Approach

Question

An object is thrown upward from the top of a 64-foot building with an initial velocity of 48 feet per second. (a) Write the position, velocity, and acceleration functions of the object. (b) When will the object hit the ground? (c) When is the velocity of the object zero? (d) How high does the object (e) Use a graphing utility to graph the position, velocity, and acceleration functions in the same viewing window. Write a short paragraph that describes the relationship among these functions.

Step-by-Step Solution

Answer

The position function of the object is \( h(t) = -16t^2 + 48t + 64 \), velocity function is \( v(t) = -32t + 48 \), and acceleration function is \( a(t) = -32 \) ft/s\(^2\). The object will hit the ground at \( t = 4 \) seconds. Velocity of the object will be zero at \( t = 1.5 \) seconds when object reaches a maximum height of 100 feet.

Step 1: Write the Position Function

The general equation in physics for vertical motion is \( h(t) = -16t^2 + v_0t + h_0 \) where \( h(t) \) is the height at time \( t \), \( v_0 \) is the initial velocity, and \( h_0 \) is the initial height. For our case, the initial velocity is 48 feet per second, and the initial height is 64 feet. Therefore, the position function becomes \( h(t) = -16t^2 + 48t + 64 \)

Step 2: Write the Velocity Function

The velocity function is the derivative of the position function, \( v(t) = h'(t) \). Derivative of \( h(t) \) gives \( v(t) = -32t + 48 \).

Step 3: Write the Acceleration Function

The acceleration function is the derivative of the velocity function, \( a(t) = v'(t) \). Since velocity function is linear, its derivative is constant, thus \( a(t) = -32 \) ft/s\(^2\). This is the acceleration due to gravity, which is constant.

Step 4: Find When Object Hits the Ground

The object hits the ground when height is zero, so solve the equation \( h(t) = 0 \) for \( t \). Solving \( -16t^2 + 48t + 64 = 0 \) using the quadratic formula gives us the solution \( t = 4 \) seconds.

Step 5: Find When Velocity is Zero

The velocity of the object will be zero at maximum height, so solve the equation \( v(t) = 0 \) for \( t \). Solving \( -32t + 48 = 0 \) for \( t \) gives us \( t = 1.5 \) seconds.

Step 6: Find Maximum Height

Substitute \( t = 1.5 \) into the height function to find maximum height, \( h(1.5) = -16(1.5)^2 + 48(1.5) + 64 \), results in \( h(1.5) = 100 \) feet. Thus, the object reaches a maximum height of 100 feet.

Step 7: Graphing and Explanation

We can clearly see from the graph that the functions are interrelated. As the object is thrown, it rises reaching a maximum height and then falls back to the ground. During the upward motion, velocity decreases and becomes zero at the highest point. Acceleration remains the same throughout because of the constant gravitational pull. Both height and velocity are parabolic while acceleration is constant.

Key Concepts

Position Function

In vertical motion problems, the position function gives us the height of an object at any time, based on its initial conditions. This is a crucial concept in understanding how objects move under the influence of gravity. For our exercise, an object is thrown from a building, and the position function is established using the formula: \[ h(t) = -16t^2 + v_0t + h_0 \] Here, \( h(t) \) represents the height at time \( t \), \( v_0 \) is the initial velocity (48 feet per second in this scenario), and \( h_0 \) is the initial height (64 feet). Thus, the position function simplifies to: \[ h(t) = -16t^2 + 48t + 64 \] This equation reflects the parabolic path of the object, accounting for the effects of gravity, which acts downward at \(-16t^2\). Possessing the position function is vital as it allows us to predict when the object will hit the ground, reach a certain height, or return to the initial height.

Velocity Function

The velocity function is the derivative of the position function, providing the speed of the object at any point in its motion. It tells us how fast and in which direction the object is moving. In mathematical terms, if the position function is \( h(t) = -16t^2 + 48t + 64 \), then the velocity function \( v(t) \) is derived as: \[ v(t) = h'(t) = -32t + 48 \]

• The velocity is positive when the object moves upward.
• The velocity becomes zero when the object is momentarily at rest at its peak height.
• Then, as the object falls, the velocity becomes negative.

This function shows that the velocity linearly decreases due to gravity. In this case, it starts at 48 feet per second and diminishes over time as gravity exerts a constant pull.

Acceleration Function

The acceleration function gives the rate of change of velocity with respect to time. For any object in vertical motion under gravity, acceleration is constant, reflecting gravity's influence. The derivative of the velocity function \( v(t) = -32t + 48 \) leads to a simple calculation for acceleration: \[ a(t) = v'(t) = -32 \] Here, \( a(t) = -32 \) feet per second squared indicates that gravity is pulling the object downwards with a constant acceleration value. This is a hallmark of gravitational force, impacting all objects equally. Understanding this, one can see how acceleration remains unchanged throughout the object's flight, steadily decreasing the velocity until it's zero, marking the top of the object's ascent.

Parabolic Motion

Parabolic motion is seen whenever an object is acted upon by gravity, such as a thrown ball or a rocket launch. The path of these objects forms a parabola due to the constant acceleration from gravity. For the given exercise involving an object thrown upward, the position and velocity functions both exemplify parabolic characteristics.

• The position function \( h(t) = -16t^2 + 48t + 64 \) describes a downward-opening parabola.
• Initially, the object ascends to its peak height and then descends.

The velocity follows suit, slowing down to zero at the peak and increasing in the negative direction as it falls. Accurately graphing these functions illuminates the relationship among them, helping to visualize the transition from increasing to decreasing velocity, all governed by the constant acceleration term.