Problem 85 - Calculus and its applications

Question

For a freely falling object, \(a(t)=-32 \mathrm{ft} / \mathrm{sec}^{2},\) \(v(0)=\) initial velocity \(=v_{0}(\) in \(\mathrm{ft} / \mathrm{sec}),\) and \(s(0)=\) initial height \(=s_{0}(\) in \(\mathrm{ft}\) ). Find a general expression for \(s(t)\) in terms of \(v_{0}\) and \(s_{0}\)

Step-by-Step Solution

Answer

The position function is \( s(t) = -16t^2 + v_0t + s_0 \).

Step 1: Define the Acceleration Function

The acceleration of the object is given by \( a(t) = -32 \ \mathrm{ft/s^2} \), representing the constant acceleration due to gravity.

Step 2: Integrate Acceleration to Get Velocity

To find the velocity function \( v(t) \), integrate the acceleration function with respect to time: \[ v(t) = \int a(t) \, dt = \int (-32) \, dt = -32t + C_1 \]Since \( v(0) = v_0 \), substitute to find \( C_1 \):\[ v(0) = -32(0) + C_1 = v_0 \Rightarrow C_1 = v_0 \]Thus, \( v(t) = -32t + v_0 \).

Step 3: Integrate Velocity to Get Position

To determine the position function \( s(t) \), integrate the velocity function: \[ s(t) = \int v(t) \, dt = \int (-32t + v_0) \, dt \]\[ = -16t^2 + v_0t + C_2 \]Given \( s(0) = s_0 \), substitute to find \( C_2 \): \[ s(0) = -16(0)^2 + v_0(0) + C_2 = s_0 \Rightarrow C_2 = s_0 \]Thus, \( s(t) = -16t^2 + v_0t + s_0 \).

Step 4: Compose the Final Expression

The final expression for the position function in terms of time \( t \), initial velocity \( v_0 \), and initial height \( s_0 \) is: \[ s(t) = -16t^2 + v_0t + s_0 \]

Key Concepts

Understanding Acceleration in Free Fall

When discussing motion, especially free fall, acceleration plays a crucial role. Acceleration describes how quickly an object's velocity changes. In the context of free fall, acceleration is constant and downward, due to gravity.
For objects near the surface of Earth, this acceleration is approximately \(-32 \, \text{ft/s}^2\). This negative sign shows that gravity acts downwards.
In simpler terms, every second, the object's velocity increases by \(32 \, \text{ft/s}\) in a downward direction.
This constant acceleration can dramatically affect the motion and position of the falling object. Understanding it is key to predicting how the object's velocity and position will change over time.

Deriving Velocity from Acceleration

Velocity indicates how fast something is moving and in which direction. In calculus, to find velocity from acceleration, we integrate the acceleration function over time.
With our exercise, given that \( a(t) = -32 \, \text{ft/s}^2\), integrating with respect to time gives us: \[ v(t) = \int (-32) \, dt = -32t + C_1 \].
The constant \( C_1 \) results from integration, representing velocity at the initial time \( t = 0 \).
Substituting the initial condition \( v(0) = v_0 \), helps us determine \( C_1 \) as \( v_0 \).
This step results in a general velocity function: \( v(t) = -32t + v_0 \), describing how velocity changes with time.

• The slope \(-32\) suggests how rapidly velocity increases in the negative direction.
• The term \( v_0 \) sets the object's starting velocity.

Position Function and Its Role in Motion

The position function \( s(t) \) tells us where an object is at any given time. To derive this, we integrate the velocity function. For our problem, integrate \( v(t) = -32t + v_0 \) over time to obtain: \[ s(t) = \int (-32t + v_0) \, dt = -16t^2 + v_0t + C_2 \].
The constant \( C_2 \) reflects the initial position \( s_0 \). Using the condition \( s(0) = s_0 \), we confirm \( C_2 = s_0 \).
The final position function \( s(t) = -16t^2 + v_0t + s_0 \) tells us how the object's position evolves:

• The coefficient \(-16t^2\) describes how gravity affects the object over time, reflecting acceleration.
• \( v_0t \) quantifies the influence of the initial velocity.
• \( s_0 \) anchors the object to its starting point.

This function predicts where an object will be after any time, helping to understand real-world motion.