Find the distance which an object moves in time \(t\) if it starts from rest and has an acceleration \(d^{2} x / d t^{2}=g e^{-k t} .\) Show that for small \(t\) the result is approximately (1.10), and for very large \(t\), the speed \(d x / d t\) is approximately constant. The constant is called the terminal speed. (This problem corresponds roughly to the motion of a parachutist.)

• Kinematics is the branch of physics that studies motion without considering the forces that cause it.
• It focuses on quantities like displacement, velocity, and acceleration.
• The fundamental equations of kinematics are used to describe the motion of objects.

The exercise involves calculating how far an object moves over time with a continuously changing acceleration. Since the object starts from rest, its initial position and velocity are zero. This ensures a straightforward integration process, connecting acceleration to velocity and then to position. Understanding how these quantities change with time is key to solving kinematic problems efficiently.

Acceleration describes how quickly an object's velocity changes. In this problem, the acceleration is given by \(\frac{d^{2} x}{d t^{2}} = g e^{-k t}\). Note that this acceleration is not constant; it decreases over time. To solve for distance, we first needed the velocity by integrating the acceleration with respect to time:
\[ v(t) = \frac{d x}{d t} = -\frac{g}{k} e^{-k t} + \frac{g}{k} \]
Starting from rest means initial velocity, \(v(0) = 0\), making our integration straightforward. Acceleration thus affects how the object speeds up or slows down, dictating its journey from starting point to a certain distance over given time intervals.

Integration is a fundamental mathematical process in physics. It allows us to find quantities like velocity and position from acceleration. In this exercise, we integrated the acceleration function to find the velocity and then integrated the velocity to determine the position.
First, we found velocity: \[\frac{d x}{d t} = v(t) = -\frac{g}{k} e^{-k t} + \frac{g}{k} \]
Next, integrating velocity to find the position:
\[ x(t) = \frac{g}{k^2}(1 - e^{-k t}) + C_2 \]
By applying initial conditions (starting from rest), we solved for any constants and simplified our expressions. Understanding and applying integration in these steps is crucial for translating how acceleration impacts velocity and, subsequently, position over time.