Problem 42 - Thomas Calculus in SI Units

Question

Give the velocity \(v=d s / d t\) and initial position of an object moving along a coordinate line. Find the object's position at time \(t\). $$v=32 t-2, \quad s(0.5)=4$$

Step-by-Step Solution

Answer

The object's position at time \( t \) is \( s(t) = 16t^2 - 2t + 1 \).

Step 1: Write down the Given Information

We are given the velocity function of an object as \( v(t) = 32t - 2 \). We are also given the initial position of the object as \( s(0.5) = 4 \). Our task is to find the position function \( s(t) \) of the object.

Step 2: Define the Relationship Between Position and Velocity

The velocity function \( v(t) \) is the derivative of the position function \( s(t) \). Therefore, we have:\[v(t) = \frac{ds}{dt}\]This means that to find \( s(t) \), we need to integrate \( v(t) \).

Step 3: Integrate the Velocity Function

To find the position function, we integrate the velocity function:\[s(t) = \int (32t - 2) \, dt\]This gives us:\[s(t) = 16t^2 - 2t + C\]where \( C \) is the constant of integration.

Step 4: Use Initial Condition to Solve for the Constant

We use the initial condition \( s(0.5) = 4 \) to find the constant \( C \). Plug the initial condition into the position function:\[4 = 16(0.5)^2 - 2(0.5) + C\]Calculate the values:\[4 = 16(0.25) - 1 + C\]and simplify:\[4 = 4 - 1 + C\]\[C = 1\]

Step 5: Write the Final Position Function

Now that we know \( C = 1 \), substitute it back into the expression for \( s(t) \):\[s(t) = 16t^2 - 2t + 1\]This is the position function for the object at time \( t \).

Key Concepts

Differential Calculus

Differential calculus is all about understanding how things change. Think of it as a tool to find rates of change, like velocity. In physics, when an object moves, we often care about how fast it’s moving or its **velocity**. The velocity is essentially the derivative of the position with respect to time.
For instance, if you're observing the position function of an object, its derivative will tell you how the position changes with time. In our example, we have a velocity function:

• \(v(t) = 32t - 2\)

This means the velocity depends on time. By taking the derivative of the position (or simply looking at the velocity function), we learn about the object's speed at any given moment.
So, differential calculus helps us convert the initial problem of finding a changing position into finding a fixed velocity at any point in time. This is the first step in figuring out the position at any given time.

Integration

Integration is like the reverse of differentiation. While differentiation breaks down functions to look at their rate of change, integration assembles these rates to recover the original function. When you have a velocity function, as we do with \(v(t) = 32t - 2\), integrating it will give you the position function, \(s(t)\).
To find the position function from the velocity, you integrate as follows:

• \(s(t) = \int (32t - 2) \, dt\)
• Resulting in: \(s(t) = 16t^2 - 2t + C\)

Here, \(C\) is a constant of integration, because an indefinite integral is involved. It's like an unknown part of the puzzle that we will solve using initial conditions. Integration allows us to fill in the gaps from information about changes to find the position function.

Initial Conditions

Initial conditions are crucial because they give us the missing piece of the puzzle when dealing with integration. When we integrate, we're left with a constant \(C\), which can take any value unless specified further. Initial conditions help us find this specific value.
Our initial condition is given as \(s(0.5) = 4\). This tells us that at time \(t = 0.5\), the position of the object is 4. To find \(C\), we plug this condition into the position equation, like this:

• \(4 = 16(0.5)^2 - 2(0.5) + C\)
• Solving gives \(C = 1\)

Substituting back, we get the final formula for the position function:
\(s(t) = 16t^2 - 2t + 1\)
By incorporating this starting condition, we tailor the general solution to the specific situation, ensuring that our position function accurately reflects the object's movement over time.