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Chapter 4: Problem 48

Give the acceleration \(a=d^{2} s / d t^{2},\) initial velocity, and initial position of an object moving on a coordinate line. Find the object's position at time \(t .\) \begin{equation}a=\frac{9}{\pi^{2}} \cos \frac{3 t}{\pi}, \quad v(0)=0, \quad s(0)=-1\end{equation}

Short Answer

Expert verified
The position function is \( s(t) = -\cos\left(\frac{3t}{\pi}\right) \).

Step by step solution

01

Integrate Acceleration to Find Velocity

The acceleration function is given by \( a(t) = \frac{9}{\pi^2} \cos \frac{3t}{\pi} \). To find the velocity function \( v(t) \), we need to integrate the acceleration function with respect to \( t \). Thus, \(\int a(t) \, dt = \int \frac{9}{\pi^2} \cos \frac{3t}{\pi} \, dt.\)Using the substitution \( u = \frac{3t}{\pi} \) gives \( du = \frac{3}{\pi} dt \) or \( dt = \frac{\pi}{3} du \). The integral becomes\( \frac{9}{\pi^2} \int \cos u \frac{\pi}{3} \, du = \frac{3}{\pi} \int \cos u \, du \).Evaluating the integral, \( \frac{3}{\pi} \sin u + C = \frac{3}{\pi}\sin\left(\frac{3t}{\pi}\right) + C \).So, \( v(t) = \frac{3}{\pi}\sin\left(\frac{3t}{\pi}\right) + C \).
02

Use Initial Velocity Condition

We have the initial velocity condition \( v(0) = 0 \). Substitute \( t = 0 \) into the velocity equation:\[v(0) = \frac{3}{\pi} \sin\left(\frac{3 \cdot 0}{\pi}\right) + C = 0.\]Since \( \sin(0) = 0 \), we find that \( C = 0 \).Thus, the velocity function is:\[v(t) = \frac{3}{\pi}\sin\left(\frac{3t}{\pi}\right).\]
03

Integrate Velocity to Find Position

To find the position function \( s(t) \), integrate the velocity function \( v(t) \). Thus,\[ s(t) = \int \frac{3}{\pi} \sin\left(\frac{3t}{\pi}\right) \, dt. \]Using the substitution \( u = \frac{3t}{\pi} \), \( du = \frac{3}{\pi} dt \), we have \( dt = \frac{\pi}{3} du \). Then the integral becomes\[\int \frac{3}{\pi} \sin u \frac{\pi}{3} \, du = \int \sin u \, du = -\cos u + C = -\cos\left(\frac{3t}{\pi}\right) + C.\]Therefore,\[ s(t) = -\cos\left(\frac{3t}{\pi}\right) + C.\]
04

Use Initial Position Condition

The initial position condition is \( s(0) = -1 \). Substitute \( t = 0 \) into the position function to find \( C \):\[s(0) = -\cos\left(\frac{3 \cdot 0}{\pi}\right) + C = -1. \]\( \cos(0) = 1 \), so\[ -1 + C = -1, \] which gives \( C = 0 \).Then the position function is:\[ s(t) = -\cos\left(\frac{3t}{\pi}\right) \].
05

Conclusion: Position at Any Time t

The position function describing the object's position over time is\[ s(t) = -\cos\left(\frac{3t}{\pi}\right). \]This position function is derived by integrating the given acceleration and applying initial conditions for velocity and position.

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Key Concepts

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

Acceleration
Understanding acceleration is crucial in solving motion problems. Acceleration is the rate of change of velocity with respect to time. In simple terms, it tells us how quickly the velocity of an object is changing. For example, if an object's velocity is increasing, it has a positive acceleration; if decreasing, a negative acceleration. Acceleration is typically represented as \( a(t) = \frac{d^2 s}{dt^2} \), where \( s \) is the position function.
In our exercise, the acceleration function is given as \( a(t) = \frac{9}{\pi^2} \cos \left(\frac{3t}{\pi}\right) \). This function indicates that the acceleration is not constant but rather oscillates over time due to the cosine component.
  • The presence of \( \cos \left(\frac{3t}{\pi}\right) \) means the acceleration follows a wave-like pattern.
  • The constant \( \frac{9}{\pi^2} \) multiplies this wave, indicating the magnitude or amplitude of the acceleration.
Understanding this behavior helps us anticipate how the velocity and position of the object will change over time.
Velocity
Velocity represents an object's speed and direction. It is the first derivative of the position function with respect to time, indicating how the position changes over time. In mathematical terms, velocity \( v(t) \) is the integral of acceleration.
In this problem, to find the velocity function, we integrated the given acceleration, resulting in \( v(t) = \frac{3}{\pi} \sin \left(\frac{3t}{\pi}\right) + C \). After applying the initial condition \( v(0) = 0 \), we discovered \( C = 0 \), simplifying our velocity function to \( v(t) = \frac{3}{\pi} \sin \left(\frac{3t}{\pi}\right) \).
  • This formula reveals that velocity also oscillates over time, with its behavior governed by a sine function.
  • The constant \( \frac{3}{\pi} \) dictates the amplitude of these oscillations.
  • The velocity changes as the argument of the sine function changes with time \( t \).
Thus, understanding velocity allows us to predict how fast and in what direction an object's position will change.
Position Function
The position function \( s(t) \) describes an object's specific location on a coordinate line at any given time. To find the position function, we integrate the velocity function.
In this particular exercise, after integrating the velocity function, we obtained \( s(t) = -\cos \left(\frac{3t}{\pi}\right) + C \). To find \( C \), we used the initial position condition \( s(0) = -1 \). This calculation showed that \( C = 0 \), which simplified the position function to \( s(t) = -\cos \left(\frac{3t}{\pi}\right) \).
  • The \( \cos \left(\frac{3t}{\pi}\right) \) term implies that the position oscillates; it moves back and forth over time as a factor of cosine function.
  • The negative sign in front signifies the direction of initial movement relative to the reference point.
  • Evaluating this function at various times \( t \), allows predicting an object's precise location on its path.
By understanding the position function, we gain insight into how the initial conditions and time influence an object's trajectory.

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Most popular questions from this chapter

Find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function's first and second derivatives. How are the values at which these graphs intersect the \(x\)-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? \begin{equation}y=x^{5}-5 x^{4}-240\end{equation}

Suppose the derivative of the function \(y=f(x)\) is \begin{equation}y^{\prime}=(x-1)^{2}(x-2).\end{equation} At what points, if any, does the graph of \(f\) have a local minimum, local maximum, or point of inflection? (Hint: Draw the sign pattern for \(y^{\prime}.)\)

Find the curve \(y=f(x)\) in the \(x y\) -plane that passes through the point \((9,4)\) and whose slope at each point is 3\(\sqrt{x} .\)

\(\begin{array}{l}{\text { Stopping a car in time You are driving along a highway at a }} \\ {\text { steady } 60 \mathrm{mph}(88 \mathrm{ft} / \mathrm{sec}) \text { when you see an accident a head and }} \\ {\text { slam on the brakes. What constant deceleration is required to }} \\ {\text { stop your car in } 242 \mathrm{ft} \text { ? To find out, carry out the following }} \\ {\text { steps. }}\end{array}\) \(\begin{array}{l}{\text { 1. Solve the initial value problem }} \\ {\text { Differential equation: } \frac{d^{2} s}{d t^{2}}=-k \quad(k \text { constant })} \\ {\text { Initial conditions: } \frac{d s}{d t}=88 \text { and } s=0 \text { when } t=0}\end{array}\) \(\begin{array}{l}{\text { 2. Find the value of } t \text { that makes } d s / d t=0 . \text { (The answer will }} \\ {\text { involve } k . )} \\ {\text { 3. Find the value of } k \text { that makes } s=242 \text { for the value of } t} \\ {\text { you found in Step } 2 \text { . }}\end{array}\)

Solve the initial value problems in Exercises \(71-90\) . $$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}, \quad y(4)=0$$
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