Question

An object moves on a line with velocity \(v(t)=\frac{10}{(t+1)^{2}} \mathrm{mi} / \mathrm{hr},\) for \(t \geq 0,\) where \(t\) is measured in hours. What is the maximum distance the object can travel?

Short answer

Explain your answer. **Answer:** The object will not reach a maximum distance for \(t\geq0\), as the position function keeps increasing without a bound, and the object is always moving forward. The velocity of the object approaches 0 as \(t\) approaches infinity, meaning the object is slowing down but never stops. Thus, the object can travel an infinite distance on the line.

Step-by-step solution

Calculate the antiderivative of the velocity function

To find the distance traveled, we need to integrate the velocity function \(v(t) = \frac{10}{(t+1)^{2}}\) with respect to time. The integral of \(\frac{10}{(t+1)^{2}}\) with respect to \(t\) can be determined by using the substitution method. Let \(u = t + 1\). Then, \(\frac{du}{dt} = 1\), and \(dt = du\). The integral becomes: $$\int{\frac{10}{u^{2}} du}$$ #Step 2: Evaluate the integral

Solve the integral

Now, solve the indefinite integral: $$\int{\frac{10}{u^{2}} du} = -\frac{10}{u} + C,$$ where \(C\) is the constant of integration. Since \(u = (t + 1)\), the expression for the distance as a function of time is: $$D(t) = -\frac{10}{t + 1} + C$$ #Step 3: Identify the position at t=0

Evaluate the position at t=0

To find the constant of integration, we can determine the position of the object at \(t=0\). Since the object starts moving at \(t=0\), its initial position, \(D(0)\), should be 0. $$D(0) = -\frac{10}{0+1} + C = 0$$ Solving for \(C\), we get \(C=10\). Thus, the position function becomes: $$D(t) = -\frac{10}{t + 1} + 10$$ #Step 4: Identify if the position function has a maximum

Determine if the position function has a maximum

To see if the position function has a maximum value for \(t\geq0\), we can examine its derivative, which is the velocity function: $$v(t)=\frac{10}{(t+1)^{2}}$$ For \(t \geq 0\), the velocity function is always positive, meaning that the object is always moving forward and its position is always increasing. As \(t\) approaches infinity, the denominator becomes larger and the velocity function approaches 0, meaning the object is slowing down but never stops, and the position function keeps increasing. Therefore, the maximum distance the object can travel is not reached, since the position function keeps increasing without a bound for \(t\geq0\). The object can travel an infinite distance on the line.

Key concepts

Antiderivative

The concept of an antiderivative, essentially, is finding a function's original form before differentiation. When faced with a function like a velocity function, discovering its antiderivative allows us to unearth another important function: the position function. This process involves taking a velocity function, like our example function, \( v(t) = \frac{10}{(t+1)^{2}} \), and integrating it over time to discover the position the object traverses.
Understanding the antiderivative is essential because when we calculate the integral of the velocity function, we add a constant of integration, commonly denoted as \( C \). This constant is crucial for determining the actual position function from any starting point. By analyzing the antiderivative, we can transition from knowing the rate of motion (velocity) to quantifying the actual distance covered by the moving object.

Velocity Function

The velocity function defines how fast an object is moving and in which direction at any given moment. In our exercise, \( v(t) = \frac{10}{(t+1)^{2}} \), the velocity function provides us with the speed of the object as a function of time \( t \).
Here, the velocity is positive, indicating forward motion along a line. This function is crucial, as it represents the rate of change of position, which ties into calculus by relating differential and integral calculus.
What we can observe from the given velocity function is that, although always positive, it decreases in magnitude as time progresses. That means the object continuously moves forward but at a decreasing rate as \( t \) increases. By studying the velocity function, we determine the behavior of the object's motion and predict how it will travel over time.

Definite Integral

The definite integral is a powerful tool to calculate the accumulated total of a quantity across an interval. In its core, think of it as summing up infinitely small pieces to find something bigger; in this case, the total distance traveled by an object.
When dealing with velocity and distance, the definite integral offers a method to measure the change over time. By integrating the velocity function over a specified time frame, we determine how far the object has traveled from the initial point to a later time.
But, in our exercise, since the time interval's upper bound is infinite, the position function calculated exhibits continuous increase, explaining why the object's position is not bounded and why it can continue to travel infinitely. This demonstrates that when evaluating a definite integral from zero to infinity, analyzing the behavior of the function becomes essential.

Distance Calculation

Distance calculation in the context of calculus integration involves finalizing the travel distance of an object through the position function. Once we find the antiderivative, and thus the position function, we can determine how far an object has moved from its starting point over any given time period.
In our example, we derived the distance function \( D(t) = -\frac{10}{t + 1} + 10 \) by integrating the velocity function. This result helps us picture how the object moves along its path. The significant part of distance calculation is specifying initial conditions. Here, the object starts at \( D(0) = 0 \), which led us to adjust our antiderivative with the constant \( C \) to exactly fit this condition.
Making these adjustments allows the distance calculation to reflect reality accurately. Despite the object's continual forward movement, and thus constant increase in position, at any theoretical moment, you can determine its position using the complete position function.