Problem 41 - Thomas Calculus

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 \(t.\) \begin{equation}v=9.8 t+5, \quad s(0)=10\end{equation}

Step-by-Step Solution

Answer

The position function is \(s(t) = 4.9t^2 + 5t + 10\).

Step 1: Set up the Position Function

We know that velocity is the derivative of position with respect to time, so we need to integrate the velocity function to find the position function. The given velocity function is \(v = \frac{d s}{d t} = 9.8t + 5\). We want to find \(s(t)\) such that \(\frac{d s}{d t} = 9.8t + 5\).

Step 2: Integrate the Velocity Function

To find \(s(t)\), integrate the velocity function with respect to \(t\): \[s(t) = \int (9.8t + 5)\, dt.\] Apply the power rule of integration for each term in the velocity function.

Step 3: Apply the Power Rule for Integration

Integrate each term separately: - For \(9.8t\), the integral is \(\frac{9.8t^2}{2} = 4.9t^2\).- For \(5\), the integral is \(5t\).Thus, the integral of the velocity function is \[s(t) = 4.9t^2 + 5t + C\] where \(C\) is the constant of integration.

Step 4: Determine the Constant of Integration

Use the initial condition \(s(0) = 10\) to find \(C\). Substitute \(t = 0\) into the position function: \[10 = 4.9(0)^2 + 5(0) + C\]\[10 = C\].So, \(C = 10\).

Step 5: Write the Position Function

With \(C\) determined, write the complete position function for the object:\[s(t) = 4.9t^2 + 5t + 10\].

Key Concepts

Velocity

In physics, velocity is an essential concept when studying the motion of objects. Essentially, velocity measures how fast something is moving and in what direction. Unlike speed, which only tells you how fast an object is moving, velocity is a vector quantity, meaning it includes both magnitude and direction.
For example, when given a velocity function like \( v(t) = 9.8t + 5 \), it describes how quickly an object is moving along a path over time. Here, the variable \( t \) represents time. The equation shows that velocity changes with time, a linear function in this case.
Velocity can change over time due to acceleration, which is the rate of change of velocity with respect to time. Acceleration can be constant or variable, affecting how the velocity function is constructed. With constant acceleration, as in our example, velocity is a linear function of time.

Integration

Integration is a fundamental concept in calculus that allows us to find quantities like area under curves or, in the context of motion, obtain a position function from a velocity function. When you integrate a velocity function, you're essentially adding up all the small changes in position to find the total position at any given time.
To integrate the velocity function \( v(t) = 9.8t + 5 \), we set up the integral \( s(t) = \int (9.8t + 5) \, dt \), where \( s(t) \) represents the position function. The integral splits into two parts:

• The term \( 9.8t \), when integrated using the power rule (\( \int t^n dt = \frac{t^{n+1}}{n+1} \)), becomes \( 4.9t^2 \).
• The constant term \( 5 \) simply integrates to \( 5t \).

Thus, the integral of the velocity function gives us: \[ s(t) = 4.9t^2 + 5t + C \] where \( C \) is the constant of integration necessary to account for initial conditions.

Initial Conditions

Initial conditions are crucial when solving problems involving differential equations, such as finding a position from a velocity function. They allow us to determine unknown constants in the equations.
In our problem, the initial condition is given as \( s(0) = 10 \), which represents the object's starting position at time \( t = 0 \). Knowing this, we can substitute it into our integrated function \( s(t) = 4.9t^2 + 5t + C \) to find value for \( C \).
By plugging in \( t = 0 \) and \( s(0) = 10 \), we solve:

• \( 10 = 4.9(0)^2 + 5(0) + C \)
• This simplifies to \( 10 = C \)

This means the constant \( C \) equals 10, ensuring our position function accounts for where the object started. This is why the final position function is \( s(t) = 4.9t^2 + 5t + 10 \). By using initial conditions, we connect theoretical math with real-world starting points.