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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 \(t\). $$a=9.8, \quad v(0)=-3, \quad s(0)=0$$

Short Answer

Expert verified
The object's position at time \(t\) is \(s(t) = 4.9t^2 - 3t\).

Step by step solution

01

Integrate the Acceleration to Find Velocity

The velocity function can be obtained by integrating the acceleration with respect to time. \[v(t) = \int a \, dt = \int 9.8 \, dt = 9.8t + C_1\]We know the initial velocity, \(v(0) = -3\). Thus, we can find \(C_1\):\[v(0) = 9.8(0) + C_1 = -3 \ C_1 = -3\] The velocity function is: \[ v(t) = 9.8t - 3 \]
02

Integrate the Velocity to Find Position

The position function can be obtained by integrating the velocity function with respect to time.\[s(t) = \int v(t) \, dt = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C_2\]We know the initial position, \(s(0) = 0\). Thus, we can find \(C_2\): \[s(0) = 4.9(0)^2 - 3(0) + C_2 = 0 \ C_2 = 0\]The position function is: \[ s(t) = 4.9t^2 - 3t \]
03

Finalize the Position Function

Now, the object's position as a function of time \(t\) is given by:\[s(t) = 4.9t^2 - 3t\]This quadratic equation describes the position of the object at any time \(t\) given the initial conditions.

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Key Concepts

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

Acceleration
When an object moves, its speed can change over time. This change in velocity per unit time is what we call acceleration. In many physics problems, like the one at hand, acceleration is a constant value. Here, the acceleration is given as 9.8 m/s虏, which often resembles the acceleration due to gravity on Earth's surface. It's essential to understand that the acceleration dictates how the velocity of the object changes over time.
  • Constant acceleration means that the increase in velocity is the same for every passing moment.
  • The formula for acceleration is given as: \[a = \frac{d^2s}{dt^2}\]where \( s \) is the position.
  • Understanding acceleration helps in predicting how quickly and in what manner an object speeds up or slows down.
Velocity Integration
To find how the position of an object changes over time, we first need to determine its velocity function. Velocity integration is the process where we integrate the given acceleration to find the velocity. In our example:\[v(t) = \int a \, dt = 9.8t + C_1\]
To find the constant \( C_1 \), we apply the initial condition provided, which is the initial velocity \( v(0) = -3 \). This step is crucial because:
  • Without the initial velocity, we wouldn't be able to find the complete and precise velocity formula.
  • Finding \( C_1 \) adjusts the equation to match the initial scenario of the problem.
The integrated velocity function helps us understand the speed of the object at any time \( t \). It's a key piece of the puzzle for later determining position.
Position Equation
Once the velocity function is established, the next step is to find the position function of the object, which represents its location at any time \( t \). This involves another step of integration:\[s(t) = \int v(t) \, dt = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C_2\]Here, the integration of the velocity gives a quadratic function, which commonly appears in problems involving constant acceleration.
  • The quadratic term \( 4.9t^2 \) results due to the constant acceleration over time.
  • The linear term \( -3t \) arises from the component of velocity that is independent of time squared.
To find \( C_2 \), we use the initial position \( s(0) = 0 \), which matches the starting condition stated in the problem.
Initial Conditions
Initial conditions are essential to fully solve the problem, as they provide the specific parameters that differentiate one scenario from another. They allow us to pin down constants from our integrations above, tailoring the generalized solution of differential equations to a particular situation.
  • Here, the initial velocity \( v(0) = -3 \) m/s tells us how fast and in which direction the object was moving initially.
  • The initial position \( s(0) = 0 \) sets the starting point for measuring the object's travel path.
By incorporating these initial conditions, the final solutions for velocity and position aren't just abstract expressions. Instead, they describe the real motion of our object starting from a known state and continuing onward through time, as governed by the uniform acceleration.

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Most popular questions from this chapter

You are to construct an open rectangular box with a square base and a volume of \(48 \mathrm{ft}^{3} .\) If material for the bottom costs \(\$ 6 / \mathrm{ft}^{2}\) and material for the sides costs \(\$ 4 / \mathrm{ft}^{2}\), what dimensions will result in the least expensive box? What is the minimum cost?

a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\).

Solve the initial value problems. $$\frac{d v}{d t}=\frac{8}{1+t^{2}}+\sec ^{2} t, \quad v(0)=1$$

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a. Find a curve \(y=f(x)\) with the following properties: i) \(\frac{d^{2} y}{d x^{2}}=6 x\) ii) Its graph passes through the point (0,1) and has a horizontal tangent line there. b. How many curves like this are there? How do you know?
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