Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

The velocity function for an object is given. Assuming that the object is at the origin at time \(t=0\), find the position at time \(t=4\). $$ v(t)=1+2 t $$

Step-by-Step Solution

Answer

The position at \( t=4 \) is 20 units.

Step 1: Understand the Problem

We are given a velocity function \( v(t) = 1 + 2t \) and need to find the position of the object at time \( t=4 \). The object starts at the origin (position = 0) when \( t=0 \).

Step 2: Recall the Relationship Between Velocity and Position

The position function \( s(t) \) can be found by integrating the velocity function \( v(t) \). This is because velocity is the derivative of position. Thus, \( s(t) = \int v(t) \, dt \).

Step 3: Integrate the Velocity Function

We integrate \( v(t) = 1 + 2t \) to find the position function \( s(t) \).\[ s(t) = \int (1 + 2t) \, dt = \int 1 \, dt + \int 2t \, dt \]\[ = t + t^2 + C \]where \( C \) is the constant of integration.

Step 4: Determine the Constant of Integration

Since the object is at the origin when \( t=0 \), we use the initial condition \( s(0) = 0 \) to determine \( C \).\[ s(0) = 0 + 0^2 + C = 0 \implies C = 0 \]

Step 5: Write the Position Function

Substitute \( C = 0 \) back into the position function. We get:\[ s(t) = t + t^2 \]

Step 6: Calculate the Position at \( t=4 \)

Substitute \( t = 4 \) into the position function:\[ s(4) = 4 + 4^2 = 4 + 16 = 20 \]

Key Concepts

Velocity Function

In calculus, understanding the velocity function is crucial when studying motion. A velocity function, often represented as \( v(t) \), describes the speed and direction of an object at any given time \( t \). It tells us how fast the object's position is changing. For example, the velocity function given is \( v(t) = 1 + 2t \). This means that as time progresses, the velocity of the object increases linearly. The velocity function allows us to predict how the object will move, providing a bridge to understanding other concepts, like the position function. Knowing the velocity function is the first step in determining the object’s position over time.

Position Function

The position function, \( s(t) \), represents the location of an object at any time \( t \). Unlike velocity, which tells you how fast something is moving, the position function tells you exactly where it is located on a one-dimensional line. To find the position function, you integrate the velocity function. This relationship exists because velocity is the derivative of position. When we integrate the velocity function \( v(t) = 1 + 2t \), we find the position function to be \( s(t) = t + t^2 + C \). The integration process reveals how the object's position changes over time based on its velocity, plus an integration constant \( C \), which accounts for the initial position of the object.

Integration

Integration is a fundamental concept in calculus that is used to calculate the area under a curve. In context to velocity and position, integration helps us transition from velocity to position. For the given problem, we integrate the velocity function \( v(t) = 1 + 2t \) to find the position function:

• First, integrate the constant \( 1 \), giving \( t \).
• Then, integrate \( 2t \), which becomes \( t^2 \).

Combining these results, we get \( s(t) = t + t^2 + C \).The constant \( C \), which is determined using initial conditions, ensures the solution fits the real-life scenario. Integration is an important tool as it traces back the accumulated change in velocity to get the displacement.

Kinematics

Kinematics is the branch of classical mechanics that describes the motion of points, bodies, and systems of bodies without considering the causes of motion. It involves variables such as velocity, position, and time. In this problem, kinematics is vividly illustrated through the association of velocity and position. With a known velocity function \( v(t) = 1 + 2t \), kinematics tells us how to find the position function \( s(t) = t + t^2 \), which describes where the object is at any time. Kinematic equations are valuable as they help predict future motion by making use of initial conditions. For instance, knowing that the object is at the origin when \( t=0 \), we can find the precise constant of integration and calculate the exact position at a given time, like \( t=4 \). This makes kinematics not only a theoretical tool but also highly practical in predicting motion.