Problem 93 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Displacement from \( t=1 \) to \( t=3 \) is always 37.2, regardless of initial position. Knowing an antiderivative suffices for finding displacement without knowing exact positions.

Step 1: Integrate the Velocity Function

The body's displacement over an interval can be found by integrating the velocity function over that interval. First, find the antiderivative of the velocity function: \( \frac{d s}{d t}=v=9.8 t-3 \). Integrate this with respect to time \( t \). The integral of the velocity \( v(t) = 9.8t - 3 \) is \( s(t) = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C \), where \( C \) is a constant of integration.

Step 2: Use Initial Condition to Determine Constant C (Case i)

Given that \( s=5 \) at \( t=0 \), substitute these values into the equation to find \( C \): \( 5 = 4.9(0)^2 - 3(0) + C \). So, \( C = 5 \). Therefore, \( s(t) = 4.9t^2 - 3t + 5 \).

Step 3: Compute Displacement from t=1 to t=3 (Case i)

With \( s(t) = 4.9t^2 - 3t + 5 \), calculate \( s(t) \) at \( t=1 \) and \( t=3 \):\[ \begin{align*} s(3) &= 4.9(3)^2 - 3(3) + 5 = 44.1 \ \ s(1) &= 4.9(1)^2 - 3(1) + 5 = 6.9 \end{align*} \] Therefore, the displacement is \( s(3) - s(1) = 44.1 - 6.9 = 37.2 \).

Step 4: Use Initial Condition to Determine Constant C (Case ii)

Given that \( s=-2 \) at \( t=0 \), substitute into the equation to find \( C \): \( -2 = 4.9(0)^2 - 3(0) + C \). So, \( C = -2 \). Therefore, \( s(t) = 4.9t^2 - 3t - 2 \).

Step 5: Compute Displacement from t=1 to t=3 (Case ii)

With \( s(t) = 4.9t^2 - 3t - 2 \), calculate \( s(t) \) at \( t=1 \) and \( t=3 \):\[ \begin{align*} s(3) &= 4.9(3)^2 - 3(3) - 2 = 37.1 \ \ s(1) &= 4.9(1)^2 - 3(1) - 2 = -0.1 \end{align*} \] Therefore, the displacement is \( s(3) - s(1) = 37.1 - (-0.1) = 37.2 \).

Step 6: Use Initial Condition to Determine Displacement (Case iii)

For the general initial condition \( s = s_0 \) at \( t=0 \), the constant is \( C = s_0 \). Thus, the displacement is computed as before but without a specific value for \( C \).\[ s(3) - s(1) = (4.9(3)^2 - 3(3) + s_{0}) - (4.9(1)^2 - 3(1) + s_{0}) = 44.1 - 6.9 = 37.2 \].

Step 7: Discuss the Consequences of Antiderivatives for Displacement (Part b)

For part b, yes, it is possible to find the displacement between two times using the antiderivative of the velocity function because displacement is the definite integral of velocity over the interval \([a, b]\), independent of the initial position \( s(t_0) \). The antiderivative provides the formula for determining displacement without needing to know the initial position.

Key Concepts

Antiderivative

The antiderivative is also known as the indefinite integral of a function. It reverse-engineers the process of differentiation, finding a function whose derivative is the given function. In the context of displacement calculation from a velocity function, the antiderivative is crucial.

To find displacement, we integrate the velocity function. Consider the velocity function given as:

• \(v(t) = 9.8t - 3\)

Integrating this function with respect to time \(t\), we obtain the antiderivative:

• \( s(t) = 4.9t^2 - 3t + C\)

Here, \(C\) is the constant of integration which we later determine using initial conditions. This antiderivative represents the position function \(s(t)\), essential for finding how far a body moves over time.

Velocity Function

The velocity function describes how fast and in which direction a body's position changes over time. The given velocity function in this context:

• \(v(t) = 9.8t - 3\)

gives the rate of change of the body's position or displacement.Using the velocity function, we can determine the body's movement over a specified time period by integrating it from one point in time to another. This integration can visualize the potential movement of a body along a path. Understanding the velocity function is vital for determining further concepts like displacement and adjusting it when initial conditions come into play.

Initial Condition

Initial conditions help determine unknown constants when dealing with indefinite integrals like antiderivatives. They provide specific values at certain points, usually starting points, which integrate into the general solution to find a precise path or position function.For instance, when we know that \(s=5\) at \(t=0\), setting these values into the general antiderivative \(s(t) = 4.9t^2 - 3t + C\) allows us to pinpoint \(C\):

• \(5 = 4.9(0)^2 - 3(0) + C\)
• \(C = 5\)

Initial conditions like this enable the transition from a broad solution to a specific, meaningful one. This process ensures that we capture the real-world scenario of the body's movement, ultimately impacting the calculation of displacement.

Definite Integral

The definite integral is a tool in calculus used for calculating the net area under a curve, which, in physic, translates to the total displacement of a moving object. By integrating the velocity function over a specified interval \(a, b\), we obtain the total change in position, known as displacement.Using a velocity function:

• \(v(t) = 9.8t - 3\)

integrated over the interval from \(t=1\) to \(t=3\), the definite integral

• \[ \int_{1}^{3} (9.8t - 3) \, dt = [4.9t^2 - 3t]\_{1}^{3} = 37.2 \]

is calculated to find the exact displacement the body experiences. This process shows that the definite integral provides a complete picture of an object's movement between two points, independent from any initial position the object might have had. This integral offers an efficient method to compute displacements using only the velocity function.