Problem 97 - Thomas Calculus in SI Units

Question

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 \(d s / d t\) 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 33.2 units. Part (b): True, displacement depends only on velocity's integral over the interval.

Step 1: Understand the Problem

We need to find the displacement of a body moving with velocity \(v = 9.8t - 3\) over different intervals, and given different initial conditions for position \(s\). Displacement is the change in position over a specified time interval, which can be found by integrating the velocity function over that interval.

Step 2: Integrate the Velocity Function

To find the displacement, we integrate the velocity function \(v(t) = 9.8t - 3\) from \(t = 1\) to \(t = 3\). The integral gives the antiderivative: \(\int (9.8t - 3) \, dt = 4.9t^2 - 3t + C\).

Step 3: Evaluate at the Time Limits for Displacement

Using the antiderivative from Step 2, compute the definite integral from \(t=1\) to \(t=3\): \[\begin{aligned}4.9t^2 - 3t \bigg|_{1}^{3} &= (4.9(3)^2 - 3(3)) - (4.9(1)^2 - 3(1)) \&= (4.9(9) - 9) - (4.9(1) - 3) \&= (44.1 - 9) - (4.9 - 3) \&= 35.1 - 1.9 \&= 33.2.\end{aligned}\]This represents the displacement over \(t=1\) to \(t=3\). It equals 33.2 units.

Step 4: Analyze Given Conditions for Position

Since displacement is calculated by integrating the velocity, it is independent of the initial position values. Thus, for each given position when \(t=0\), displacement over a time interval remains consistent if the velocity function doesn't change.

Step 5: Address Part B - Displacement Independent of Initial Position

When you know an antiderivative of the velocity function, you can compute displacement over a specific interval without the need for the body's initial or final exact position. This is because displacement is solely dependent on the area under the velocity curve between those times.

Key Concepts

Velocity Function

In physics, the velocity function describes the rate of change of position with respect to time. It gives us insight into how fast and in what direction an object is moving at any given moment. For our problem, the velocity function is given by \[ v(t) = 9.8t - 3 \].This function tells us that velocity changes over time. Here, the term \(9.8t\) suggests that velocity increases linearly with time, while the constant \(-3\) adjusts this value downward.

• The initial velocity at \(t=0\) would be \(-3\), meaning the object moves backwards initially.
• As time increases, the velocity increases, eventually becoming positive, indicating forward movement.

Understanding a velocity function is critical when finding total displacement because it reflects the movement landscape over the period of interest.

Antiderivative

The antiderivative, also known as the indefinite integral, plays a key role in solving problems involving velocity and displacement. By finding the antiderivative of a velocity function, we effectively reverse the process of differentiation. This allows us to move from a rate of movement back to a position function.In our example, the antiderivative of \(v(t) = 9.8t - 3\) is found by integrating:\[\int (9.8t - 3)\, dt = 4.9t^2 - 3t + C\]Here, \(4.9t^2\) results from integrating \(9.8t\), and \(-3t\) from the constant \(-3\). The \(+ C\) represents the constant of integration, which is used to determine a specific solution given an initial condition.

• Calculating the antiderivative is a crucial first step in finding how the whole function accumulates over time.
• The antiderivative provides the foundation on which we evaluate displacement through definite integration.

Definite Integral

The definite integral is used to compute the exact displacement over a given time interval. By evaluating the antiderivative at two time limits, we find the change in position between these points. For our specific case, the definite integral of the velocity function \(v(t) = 9.8t - 3\) between \(t = 1\) and \(t = 3\) is:\[4.9t^2 - 3t \bigg|_{1}^{3} = (4.9(3)^2 - 3(3)) - (4.9(1)^2 - 3(1)) = 33.2\]This tells us that the displacement, or the net change in position, of the object from \(t=1\) to \(t=3\) is 33.2 units.

• The definite integral essentially sums up the area under the velocity function curve between two points.
• This value represents the total effective distance travelled over the interval, including changes in direction.

Understanding this process highlights how integration connects velocity with changes in position.

Position Function

A position function gives the location of a moving object at any point in time. It is often derived from the antiderivative of the velocity function by determining the specific value of the integration constant \(C\) using initial conditions. Our general antiderivative was \[ s(t) = 4.9t^2 - 3t + C \].By substituting an initial condition, such as \(s(0)=5\), we can solve for \(C\) and find a specific position function.

• For instance, with \(s(0)=5\), solving \(5 = 4.9(0)^2 - 3(0) + C\) results in \(C=5\).
• This yields the specific position function \(s(t) = 4.9t^2 - 3t + 5\).

With the position function known, we can evaluate the object's precise position at any given time. This shows how the concepts of velocity, antiderivative, and integration interlink to determine both the displacement and absolute position over time.