Problem 61
Question
Consider a particle moving along the \(x\) -axis where \(x(t)\) is the position of the particle at time \(t, x^{\prime}(t)\) is its velocity, and \(x^{\prime \prime}(t)\) is its acceleration. A particle moves along the \(x\) -axis at a velocity of \(v(t)=1 / \sqrt{t}\) \(t>0\). At time \(t=1\), its position is \(x=4\). Find the acceleration and position functions for the particle.
Step-by-Step Solution
Verified Answer
The acceleration function is \(a(t) = -1 / 2t^{3/2}\), and the position function is \(x(t) = 2t^{1 / 2} + 2\).
1Step 1: Find the acceleration function
The acceleration of the particle is the derivative of the velocity function. In this case, the given velocity function is \(v(t) = 1 / \sqrt{t}\). It can be rewritten as \(v(t) = t^{-1 / 2}\). Then, using the power rule for differentiation, where the derivative of \(t^n\) is \(n t^{n - 1}\), we can find the acceleration function \(a(t) = x''(t) = dv/dt = -1 / 2 * t^{-3/2}\) or \(a(t) = -1 / 2t^{3/2}\).
2Step 2: Find the indefinite integral of the velocity function
The position function \(x(t)\) is the integral of the velocity function \(v(t)\). We need to find the antiderivative of the velocity function, which is given by \(\int v(t) dt = \int t^{-1 / 2} dt\). The solution to this integral, according to the power rule for integration, is \(2t^{1 / 2}\).
3Step 3: Solve for position function
To solve for the position function \(x(t)\), we add an arbitrary constant \(C\), giving us \(x(t) = 2t^{1 / 2} + C\). To find \(C\), we use the given condition that at \(t = 1\), \(x = 4\). Substituting these values into the equation gives us \(C = 2\). So the position function is \(x(t) = 2t^{1 / 2} + 2\).
Key Concepts
Velocity FunctionAcceleration FunctionPosition FunctionDifferentiationIntegration
Velocity Function
A velocity function describes how fast an object is moving along a path over time. In calculus, it is often represented as a function of time, denoted by \( v(t) \).
For our particle moving along the \( x \)-axis, its velocity is expressed as \( v(t) = \frac{1}{\sqrt{t}} \). This function shows that the particle's speed decreases as time progresses.
Velocity functions are vital for understanding motion as they directly relate to the rate of change of position. When interpreting \( v(t) \), remember that a positive value indicates forward motion, while a negative value suggests the particle is moving backwards.
Calculating velocity at different times helps in predicting where an object will be in the future.
For our particle moving along the \( x \)-axis, its velocity is expressed as \( v(t) = \frac{1}{\sqrt{t}} \). This function shows that the particle's speed decreases as time progresses.
Velocity functions are vital for understanding motion as they directly relate to the rate of change of position. When interpreting \( v(t) \), remember that a positive value indicates forward motion, while a negative value suggests the particle is moving backwards.
Calculating velocity at different times helps in predicting where an object will be in the future.
Acceleration Function
The acceleration function describes how the velocity of an object changes over time. If velocity is the speed, then acceleration is how that speed changes. It's essentially the derivative of the velocity function, represented as \( a(t) \).
For the given problem, the velocity function \( v(t) = t^{-1/2} \) has an acceleration function found by differentiating this velocity function. By applying the power rule, we get \( a(t) = -\frac{1}{2} t^{-3/2} \).
This negative sign tells us that the velocity is decreasing over time, implying that the particle is slowing down. Understanding acceleration helps us comprehend changes in motion, such as speeding up or slowing down.
For the given problem, the velocity function \( v(t) = t^{-1/2} \) has an acceleration function found by differentiating this velocity function. By applying the power rule, we get \( a(t) = -\frac{1}{2} t^{-3/2} \).
This negative sign tells us that the velocity is decreasing over time, implying that the particle is slowing down. Understanding acceleration helps us comprehend changes in motion, such as speeding up or slowing down.
Position Function
The position function provides the location of an object at any given time. It's closely related to the velocity function as the integral or antiderivative of the velocity function.
In our example, the particle's position is found by integrating the velocity function \( v(t) = t^{-1/2} \). The indefinite integral gives us \( \int t^{-1/2} \, dt = 2t^{1/2} \), which represents the position function except for the constant \( C \).
To find the exact position function \( x(t) \), we need initial conditions to solve for \( C \). Here, we are told that when \( t = 1 \), \( x = 4 \). Substituting these values, we find \( C = 2 \). So, the position function is \( x(t) = 2t^{1/2} + 2 \).
In our example, the particle's position is found by integrating the velocity function \( v(t) = t^{-1/2} \). The indefinite integral gives us \( \int t^{-1/2} \, dt = 2t^{1/2} \), which represents the position function except for the constant \( C \).
To find the exact position function \( x(t) \), we need initial conditions to solve for \( C \). Here, we are told that when \( t = 1 \), \( x = 4 \). Substituting these values, we find \( C = 2 \). So, the position function is \( x(t) = 2t^{1/2} + 2 \).
Differentiation
Differentiation is a calculus technique used to find the instantaneous rate of change of a function. It's often how we calculate things like velocity or acceleration from a position function.
For our velocity function, \( v(t) = t^{-1/2} \), applying differentiation helps us find the acceleration. Using the power rule, we differentiate \( t^n \) to get \( n t^{n-1} \). This leads to an acceleration function \( a(t) = -\frac{1}{2} t^{-3/2} \).
Differentiation essentially helps break down complex motion into something more understandable, telling us how fast velocity is changing at any given moment.
For our velocity function, \( v(t) = t^{-1/2} \), applying differentiation helps us find the acceleration. Using the power rule, we differentiate \( t^n \) to get \( n t^{n-1} \). This leads to an acceleration function \( a(t) = -\frac{1}{2} t^{-3/2} \).
Differentiation essentially helps break down complex motion into something more understandable, telling us how fast velocity is changing at any given moment.
Integration
Integration is the reverse process of differentiation and helps in finding the total accumulation of change. It's used to transition from velocity to position, determining how far an object has moved from a start point.
For our velocity function, integrating \( v(t) = t^{-1/2} \) gives us the position function. The indefinite integral \( \int t^{-1/2} \, dt = 2t^{1/2} \) informs us about how position changes over time.
Integration not only helps in finding such functions but also in solving real-world problems where quantities accumulate over time, such as distance traveled or area under a curve.
For our velocity function, integrating \( v(t) = t^{-1/2} \) gives us the position function. The indefinite integral \( \int t^{-1/2} \, dt = 2t^{1/2} \) informs us about how position changes over time.
Integration not only helps in finding such functions but also in solving real-world problems where quantities accumulate over time, such as distance traveled or area under a curve.
Other exercises in this chapter
Problem 61
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