Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

The velocity of an object is $$ v(t)= \begin{cases}5 & \text { if } 0 \leq t \leq 100 \\ 6-t / 100 & \text { if } 100700\end{cases} $$ (a) Assuming that the object is at the origin at time 0 , find a formula for its position at time \(t(t \geq 0)\). (b) What is the farthest to the right of the origin that this object ever gets? (c) When, if ever, does the object return to the origin?

Step-by-Step Solution

Answer

(a) Position: piecewise function derived for each interval. (b) Farthest right at 1750. (c) Returns to origin at t = 2450.

Step 1: Determine Position from Velocity

To find the position function, integrate the velocity function. Start by finding the antiderivative for each piece of the piecewise function for velocity.1. For \(0 \leq t \leq 100\), integrate \(v(t) = 5\): \[ x(t) = \int 5 \, dt = 5t + C_1 \] Since the object starts at the origin, \(x(0) = 0\), so \(C_1 = 0\). Thus, \(x(t) = 5t\) for \(0 \leq t \leq 100\).2. For \(100 < t \leq 700\), integrate \(v(t) = 6 - t/100\): \[ x(t) = \int (6 - t/100) \, dt = 6t - \frac{t^2}{200} + C_2 \] At \(t=100\), from the first segment: \(x(100) = 500\). Therefore, \(500 = 600 - \frac{100^2}{200} + C_2\), solving gives \( C_2 = -50\). So, \(x(t) = 6t - \frac{t^2}{200} - 50\) for \(100 < t \leq 700\).3. For \(t > 700\), integrate \(v(t) = -1\): \[ x(t) = \int -1 \, dt = -t + C_3 \] At \(t=700\), from the second segment, \(x(700) = 1750\). Therefore, \(1750 = -700 + C_3\), solving gives \(C_3 = 2450\). So, \(x(t) = -t + 2450\) for \(t > 700\).

Step 2: Find the Farthest Right Point

To find the farthest right point, evaluate the position at the endpoint of each segment and any critical points in those intervals.1. Evaluate \(x(t)\) at endpoints: - At \(t=100\), \(x(100) = 500\). - At \(t=700\), \(x(700) = 1750\). - For \(t > 700\), since the position decreases ( \(v(t) = -1\)), \(x\) will not be farther right than \(x(700) = 1750\).2. The maximum position is at \(t=700\) with \(x(700) = 1750\).

Step 3: Determine When the Object Returns to the Origin

To find when the object returns to the origin, solve for \(x(t) = 0\) in each interval.1. For \(0 \leq t \leq 100\): - \(5t = 0\) gives \(t = 0\).2. For \(100 < t \leq 700\): - \(6t - \frac{t^2}{200} - 50 = 0\) simplifies to: \(t^2/200 - 6t + 50 = 0\). Solving this quadratic equation may yield a solution, but it won't be within the valid interval.3. For \(t > 700\): - \(-t + 2450 = 0\) gives \(t = 2450\). Thus, the object returns to the origin at \(t = 2450\).

Key Concepts

Understanding the Velocity Function

Let's dive into how velocity functions play a crucial role in determining an object's movement over time. A piecewise function can express different velocities at distinct time intervals. This is useful when an object's speed changes due to various conditions. For example, our case involves a velocity function divided into three segments: constant, linearly decreasing, and negative constant velocities.

• **From 0 to 100 seconds,** the object is moving at a constant speed of 5 units per second.
• **Between 100 and 700 seconds,** the velocity decreases linearly, starting from 6 units per second and decreasing by 0.01 unit per second because of the term \(\frac{t}{100}\).
• **Beyond 700 seconds,** the velocity becomes a constant \(-1\), indicating the object is moving in the opposite direction.

Understanding these segments is vital for calculating the object's position at any given time and ultimately predicting its path. Each segment affects the position formula differently, which can be determined through integration.

The Role of Integration in Calculus

Integration in calculus allows us to find the position of an object when given its velocity function. By integrating the velocity function, we derive the position function. This process is essentially finding the antiderivative of the given piecewise sections of velocity.
Here's how it works in our specific example:

• **For 0 to 100 seconds,** integrating a constant velocity \(5\) results in a position formula of \(x(t) = 5t\), where \(C_1 = 0\) because the object starts at the origin.
• **From 100 to 700 seconds,** integrating the linear segment \(6 - \frac{t}{100}\) yields \(x(t) = 6t - \frac{t^2}{200} - 50\). Here, \(C_2 = -50\) is determined from continuity at the ends of this segment.
• **After 700 seconds,** integrating the constant \(-1\) velocity gives \(x(t) = -t + 2450\), with \(C_3\) derived from continuity again, ensuring a smooth function transition between segments.

The integration process here facilitates transitioning from velocity to finding a tangible position, which provides insights into the object's journey over time.

Solving with Quadratic Equations

Quadratic equations arise when determining when an object returns to a specific position, such as the origin. In our example, for the time interval from 100 to 700 seconds, the position function \(x(t) = 6t - \frac{t^2}{200} - 50\) steps into a quadratic form when set to zero: \(\frac{t^2}{200} - 6t + 50 = 0\).

• **Using the quadratic formula,** \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = \frac{1}{200}\), \(b = -6\), and \(c = 50\), we solve for \(t\).
• However, it’s crucial to determine if the solution actually falls within the valid interval (100 to 700 seconds). Many variables can affect this outcome, excluding irrelevant solutions.

Additionally, discovering these solutions helps reveal significant milestones in the object's trajectory, such as when it shifts directions or reaches specific points in its path.