91Ó°ÊÓ

Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the \(s\) -axis is \begin{equation} \begin{array}{c}{\text { a. Suppose that the velocity of a body moving along the } s \text { -axis is }} \\ {\frac{d s}{d t}=v=9.8 t-3} \\ {\text { i) Find the body's displacement over the time interval from }} \\ {t=1 \text { to } t=3 \text { given that } s=5 \text { when } t=0 \text { . }}\\\\{\text { ii) Find the body's displacement from } t=1 \text { to } t=3 \text { given }} \\\ {\text { that } s=-2 \text { when } t=0 \text { . }} \\ {\text { iii) Now find the body's displacement from } t=1 \text { to } t=3} \\ {\text { given that } s=s_{0} \text { when } t=0 \text { . }}\\\\{\text { b. Suppose that the position } s \text { of a body moving along a coordinate }} \\ {\text { line is a differentiable function of time } t . \text { Is it true that }} \\\ {\text { once you know an antiderivative of the velocity function }}\\\\{d s / d t \text { you can find the body's displacement from } t=a \text { to }} \\\ {t=b \text { even if you do not know the body's exact position at }} \\\ {\text { either of those times? Give reasons for your answer. }}\end{array} \end{equation}

Step by step solution

01

Integrate the velocity function

Given the velocity function \( v = \frac{ds}{dt} = 9.8t - 3 \), we first need to find the antiderivative to determine the position function, \( s(t) \). This involves integrating the velocity function with respect to \( t \).\[s(t) = \int (9.8t - 3) \, dt = 9.8 \frac{t^2}{2} - 3t + C\]where \( C \) is the constant of integration.

02

Use initial condition to find C for i)

For part i), we are given \( s = 5 \) when \( t = 0 \). Substitute these values into the position function to find \( C \).\[5 = 9.8 \frac{0^2}{2} - 3(0) + C \]This simplifies to \( C = 5 \). Thus, the position function is:\[s(t) = 4.9t^2 - 3t + 5\]

03

Calculate displacement for i)

Compute the displacement from \( t = 1 \) to \( t = 3 \) using the position function for part i).\[\Delta s = s(3) - s(1)\]First calculate \( s(3) = 4.9(3)^2 - 3(3) + 5 = 44.1 - 9 + 5 = 40.1 \).\[s(1) = 4.9(1)^2 - 3(1) + 5 = 4.9 - 3 + 5 = 6.9\]Therefore, the displacement is \( 40.1 - 6.9 = 33.2 \).

04

Use initial condition to find C for ii)

For part ii), \( s = -2 \) when \( t = 0 \). Replace these values in the position function.\[-2 = 9.8 \frac{0^2}{2} - 3(0) + C \]Simplifies to \( C = -2 \). Therefore, the position function becomes:\[s(t) = 4.9t^2 - 3t - 2\]

05

Calculate displacement for ii)

For part ii), calculate the displacement from \( t = 1 \) to \( t = 3 \).\[\Delta s = s(3) - s(1)\]First calculate \( s(3) = 4.9(3)^2 - 3(3) - 2 = 44.1 - 9 - 2 = 33.1 \).\[s(1) = 4.9(1)^2 - 3(1) - 2 = 4.9 - 3 - 2 = -0.1\]Therefore, the displacement is \( 33.1 + 0.1 = 33.2 \).

06

Calculate displacement for iii)

For part iii), we don't need to know the initial position to find displacement. Recalculate only the difference using the velocity's antiderivative.\[\Delta s = \left(4.9(3)^2 - 3(3) + C\right) - \left(4.9(1)^2 - 3(1) + C\right)\]The \( C \)s cancel out, and the calculation becomes:\[\Delta s = (44.1 - 9) - (4.9 - 3) = 33.2 \]The displacement is still \( 33.2 \).

07

Understand conceptual question in part b

In part b, we discuss whether knowing an antiderivative of the velocity function is sufficient for finding displacement between two time intervals. The displacement from time \( t = a \) to \( t = b \) can be found using the antiderivative of the velocity regardless of unknown positions, as displacement is dependent on the change in position (\( s(b) - s(a) \)), not the exact starting and ending positions.

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Key Concepts

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

Antiderivative

An antiderivative is essentially the reverse of taking a derivative. In simpler terms, if you have a function and know its rate of change (or derivative), finding the antiderivative lets you determine the original function itself.
For displacement problems, given a velocity function like \( v = 9.8t - 3 \), we find the position function \( s(t) \) by integrating the velocity. This process is the act of "undoing" the derivative.

• Start with the velocity function \( v = \frac{ds}{dt} \).
• Integrate \( v \) to get \( s(t) = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C \).
• "C" represents the initial condition or the constant of integration, without which we would not get the specific position function.

Recognizing the antiderivative helps bridge the gap between rate of change (velocity) and actual position.

Velocity Function

The velocity function gives us information about how fast an object is moving and in what direction. It's a derivative of the position with respect to time. In the problem, the velocity function is given as \( v = 9.8t - 3 \).
To understand it better:

• The term \( 9.8t \) signifies acceleration; as time increases, the velocity increases linearly.
• The term \(-3\) is the initial velocity offset.

This velocity function indicates that the body is speeding up over time, and the negative component suggests an initial motion in the opposite direction. Integrating this function allows us to find the position or displacement.

Initial Conditions

Initial conditions are crucial in problems involving antiderivatives because they allow us to solve for the constant of integration \( C \) and find the particular solution.
For instance, if we know that when \( t = 0 \), \( s = 5 \), integrating the velocity function gives \[ s(t) = 4.9t^2 - 3t + C \]. Given \( s(0) = 5 \), we substitute to find \( C \):\[5 = 4.9(0)^2 - 3(0) + C \Rightarrow C = 5 \]Initial conditions tailor the general solution to the specific scenario described, enabling accurate displacement calculations. Without this, we'd only have a general idea of the change in position.

Integration in Calculus

Integration in calculus is a fundamental tool used to find quantities where rates of change are known, such as displacement from velocity.
By integrating the velocity function \( v = 9.8t - 3 \), we get the position function:

• \( s(t) = \int v \, dt = 4.9t^2 - 3t + C \)

Using integration here allows us to accumulate the total displacement over a specific period.
Importantly, integration provides the accumulated output by considering all instantaneous changes over time, which is crucial when finding total displacement from velocity functions. This method underscores the integral's role in summing or "integrating" parts into a whole.