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Chapter 4: Problem 42

Give the velocity \(v=d s / d t\) and initial position of an object moving along a coordinate line. Find the object's position at time \(t.\) \begin{equation}v=32 t-2, \quad s(0.5)=4\end{equation}

Short Answer

Expert verified
The position function is \( s(t) = 16t^2 - 2t + 1 \).

Step by step solution

01

Understand the Given Function

The velocity function, given as \( v = \frac{ds}{dt} = 32t - 2 \), describes how position \( s \) changes with time \( t \). Our goal is to find the position function \( s(t) \).
02

Set Up the Position Function

To find the position \( s(t) \), integrate the velocity function \( v(t) = 32t - 2 \). The position function can be expressed as \( s(t) = \int (32t - 2) \, dt \).
03

Perform the Integration

Integrate the velocity function: \[ s(t) = \int (32t - 2) \, dt = 16t^2 - 2t + C \]. Here, \( C \) is the integration constant that we need to determine.
04

Use Initial Condition to Find the Constant

The initial condition given is \( s(0.5) = 4 \). Substitute this into the integrated function to determine \( C \): \[ 4 = 16(0.5)^2 - 2(0.5) + C \]. Simplifying, we get \[ 4 = 4 - 1 + C \] which implies \( C = 1 \).
05

Write the Final Position Function

Substitute \( C = 1 \) back into the position function: \[ s(t) = 16t^2 - 2t + 1 \]. This is the function describing the position of the object at any time \( t \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity Function
In calculus, the velocity function is a critical concept that helps us understand how an object's position changes over time. Velocity is essentially the rate at which an object moves in a specific direction. It's a derivative of the position function, indicating how quickly the position is changing at any moment.
  • The velocity function can be expressed as \(v = \frac{ds}{dt}\), where \(s\) is the position of the object over time \(t\).
  • In the given exercise, the specific velocity function provided is \(v = 32t - 2\).
  • This tells us that the object's rate of change in position depends linearly on time.
Using a velocity function helps predict future positions by understanding past movements.
Position Function
The position function, often denoted as \(s(t)\), represents the location of an object at any given time \(t\). This function is crucial for tracking movement and making predictions about an object's future position based on its current velocity.
The position function can be derived through integration of the velocity function. Starting with the velocity \(v(t) = 32t - 2\), the task is to determine \(s(t)\). This will give us a clear picture of where the object has been and where it might be headed.
  • To find \(s(t)\), we integrate the velocity function.
  • This leads us to the integral equation \(s(t) = \int (32t - 2) \, dt\).
  • After integration, the result is \(s(t) = 16t^2 - 2t + C\), where \(C\) is a constant determined by the initial conditions.
By establishing the position function, we effectively translate velocity into spatial progress over time.
Integration
Integration is a core tool in calculus used to find quantities that accumulate over time, like area under a curve or, in this case, position from velocity.
It works as a reverse operation to differentiation, moving back from the rate of change (such as velocity) to the actual quantity (such as position).
  • In this scenario, integrating \(v(t) = 32t - 2\) aims at finding \(s(t)\).
  • The integration yields \(s(t) = 16t^2 - 2t + C\), with \(C\) being an undetermined constant.
  • Integration is essential because it allows us to sum up all the small pieces of movement (velocity) over time into a whole (position).
This approach broadens our understanding of how dynamic quantities evolve in real-world scenarios.
Initial Condition
The initial condition provides specific information to help determine unknown constants in an integrated function, ensuring unique and accurate solutions. In many cases, integration generates solutions with one or more constants, represented as \(C\), due to the indefinite nature of integration.
In this particular exercise, the initial condition is \(s(0.5) = 4\), which means at time \(t = 0.5\), the object is positioned at \(s = 4\). This crucial piece of information is used to determine the constant \(C\) in the position function:
  • Substitute the initial condition into the integrated equation: \(4 = 16(0.5)^2 - 2(0.5) + C\).
  • Simplify to find \(C = 1\).
  • Finally, incorporate \(C\) back into the position function to finalize it: \(s(t) = 16t^2 - 2t + 1\).
Thus, the initial condition guarantees our position function precisely reflects the object's movement from a known starting point.

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Most popular questions from this chapter

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