Problem 62
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, initially at rest, moves along the \(x\) -axis such that its acceleration at time \(t>0\) is given by \(a(t)=\cos t\). At the time \(t=0\), its position is \(x=3\). (a) Find the velocity and position functions for the particle. (b) Find the values of \(t\) for which the particle is at rest.
Step-by-Step Solution
Verified Answer
The velocity function is \(v(t) = \sin(t)\), the position function is \(x(t) = -\cos(t) + 4\), and the particle is at rest for \(t = n\pi\) where \(n\) is an integer.
1Step 1: Integration of Acceleration to Get Velocity
Since the velocity is the integral of the acceleration, we will perform the integral \(\int a(t) dt\). So, \(v(t) = \int\cos(t) dt = \sin(t) + C\), where \(C\) is the constant of integration. Since the initial velocity is 0, that is \(v(0) = 0\), from \(v(t)\) equation we can find that \(C = 0\). So, the velocity function is \(v(t) = \sin(t)\).
2Step 2: Integration of Velocity to Get Position
The position is the integral of the velocity, so \(x(t) = \int v(t) dt = \int \sin(t) dt = -\cos(t) + D\), where \(D\) is the constant of integration. We know that the initial position is \(x(0) = 3\), so we substitute \(t = 0\) into \(x(t)\) to solve for \(D = 4\). Thus, the position function is \(x(t) = -\cos(t) + 4\).
3Step 3: Solve When Particle is at Rest
The particle is at rest when the velocity function \(v(t) = 0\). So, we solve for \(t\) in the equation \(\sin(t) = 0\), which gives us \(t = n\pi\), where \(n\) is an integer.
Key Concepts
Understanding the Position FunctionExploring the Velocity FunctionAnalyzing the Acceleration Function
Understanding the Position Function
The position function, denoted as \(x(t)\), describes the location of a particle on the x-axis at a specific time \(t\).
This function is crucial as it tells us where the particle is located in relation to a fixed point, usually the origin. For the given problem, the position function was determined by integrating the velocity function. The initial condition given was \(x(0) = 3\), which indicates that at time \(t=0\), our particle's location on the x-axis is 3 units from the origin.To find the position function, the velocity function \(v(t)\) is integrated. By solving this, the position function becomes \(x(t) = -\cos(t) + 4\). Here, \(-\cos(t)\) comes from integrating \(\sin(t)\), and the \(+4\) is derived from the initial position.
This function is crucial as it tells us where the particle is located in relation to a fixed point, usually the origin. For the given problem, the position function was determined by integrating the velocity function. The initial condition given was \(x(0) = 3\), which indicates that at time \(t=0\), our particle's location on the x-axis is 3 units from the origin.To find the position function, the velocity function \(v(t)\) is integrated. By solving this, the position function becomes \(x(t) = -\cos(t) + 4\). Here, \(-\cos(t)\) comes from integrating \(\sin(t)\), and the \(+4\) is derived from the initial position.
- Position at \(t=0\) allows us to solve for any constants in the position function.
- The particle's position provides insight into displacement and movement patterns over time.
Exploring the Velocity Function
The velocity function, denoted as \(v(t)\), represents the rate of change of the position with respect to time. It effectively describes the particle's speed and direction along the x-axis.
In essence, it is the derivative of the position function \(x(t)\).Given that acceleration \(a(t) = \cos(t)\), the velocity function is found by integrating the acceleration. Here, we perform \(v(t) = \int \cos(t) dt = \sin(t) + C\), where \(C\) is the constant of integration determined using initial conditions.
Since the particle is initially at rest, \(v(0) = 0\) and solving this gives \(C = 0\). Hence, the velocity function simplifies to \(v(t) = \sin(t)\).
In essence, it is the derivative of the position function \(x(t)\).Given that acceleration \(a(t) = \cos(t)\), the velocity function is found by integrating the acceleration. Here, we perform \(v(t) = \int \cos(t) dt = \sin(t) + C\), where \(C\) is the constant of integration determined using initial conditions.
Since the particle is initially at rest, \(v(0) = 0\) and solving this gives \(C = 0\). Hence, the velocity function simplifies to \(v(t) = \sin(t)\).
- Initial Condition: Using \(v(0)=0\) helps in identifying constants accurately.
- The function \(v(t) = \sin(t)\) indicates oscillating motion, typical of certain periodic movements.
Analyzing the Acceleration Function
The acceleration function \(a(t)\), describes the change in velocity over time. It plays a critical role in understanding how quickly or slowly a particle's velocity changes. In our problem, the acceleration is given directly by \(a(t) = \cos(t)\). This tells us that acceleration varies with the cosine of \(t\), illustrating a wave-like pattern which can impact how the particle moves over time. Understanding acceleration is key to predicting motion paths:
- It affects the velocity, as acceleration is its derivative.
- Oscillating acceleration patterns, as indicated by the cosine function, suggest regular periodic motion changes.
Other exercises in this chapter
Problem 62
Find the particular solution of the differential equation that satisfies the initial conditions. $$ f^{\prime \prime}(x)=\sin x+e^{2 x}, \quad f(0)=\frac{1}{4},
View solution Problem 62
State the integration formula you would use to perform the integration. Do not integrate. $$ \int \frac{\sec ^{2} x}{\tan x} d x $$
View solution Problem 63
In Exercises \(63-68,\) find the limit. \(\lim _{x \rightarrow \infty} \sinh x\)
View solution Problem 63
Verify each rule by differentiating. Let \(a>0\). $$ \int \frac{d u}{u \sqrt{u^{2}-a^{2}}}=\frac{1}{a} \operatorname{arcsec} \frac{|u|}{a}+C $$
View solution