Problem 62 - Calculus

Question

The velocity of an object is $$ v(t)=\left\\{\begin{array}{ll} 5 & \text { if } 0 \leq t \leq 100 \\ 6-t / 100 & \text { if } 100700 \end{array}\right. $$ (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

The object reaches its furthest right at \( t=600 \) with position 800. It returns to the origin at \( t=800 \).

Step 1: Understanding the Velocity Function

The given velocity function is piecewise, divided into three intervals for time \( t \). The velocity is constant (5) for \( 0 \leq t \leq 100 \), decreases linearly from 6 to -1 over \( 100 < t \leq 700 \), and remains constant at -1 for \( t > 700 \).

Step 2: Finding Position Function for Each Interval

Position is the integral of velocity. For \( 0 \leq t \leq 100 \), integrate \( v(t) = 5 \) to get position \( x(t) = 5t + C \). With \( x(0) = 0 \), \( C = 0 \), so \( x(t) = 5t \).

Step 3: Position for Second Interval

For \( 100 < t \leq 700 \), integrate \( v(t) = 6 - \frac{t}{100} \). First, find constant of integration. At \( t=100 \), position is \( x(100) = 5(100) = 500 \). Integrate to get: \[ x(t) = 500 + (6t - \frac{t^2}{200}) + C \]. Solve for \( C \) using \( x(100) = 500 \).

Step 4: Position for Third Interval

For \( t > 700 \), integrate \( v(t) = -1 \). At \( t=700 \), calculate position from previous interval. Use initial condition for segment, \( x(700) \), to find position equation for \( t > 700 \). It is \( x(t) = \text{position at 700} - (t - 700) \).

Step 5: Calculate the Maximum Position to the Right

The object moves to the right when velocity is positive and starts decreasing when velocity becomes negative. Calculate the maximum position reached in the decelarating phase (from \( 100 < t \leq 700 \)). Solve for maximum position by finding where derivative of the position (velocity) equals zero.

Step 6: Solve for Return to the Origin

Set the position function from different time intervals equal to zero and solve for \( t \). Check each interval. From third interval \( t > 700 \), the object moves left; solve for when position equals zero to find return time, if possible.

Key Concepts

Velocity Function

In mathematics, the **velocity function** describes the rate at which an object changes its position over time. It's essentially the derivative of the position function with respect to time. For instance, when observing a physics problem, the velocity function can help determine how fast an object is moving and in what direction. This can be particularly interesting when dealing with non-constant velocities, as you can analyze how the velocity changes over different intervals.

In the given exercise, the velocity function is defined using a piecewise expression, which outlines the behavior of an object's speed at various stages:

• From time 0 to 100, the velocity is constant at 5, meaning the object moves steadily at a consistent rate.
• Between time 100 and 700, the velocity decreases linearly, indicating a gradual slowing down of the object.
• After time 700, the velocity is -1, suggesting the object moves back, returning toward its starting point.

Understanding these intervals allows one to construct the position function from each section of the velocity function by integration. This detailed breakdown can enrich your understanding of how velocity transitions affect the position.

Piecewise Function

A **piecewise function** is a function defined by multiple sub-functions, each applicable to a certain interval of the domain. They are particularly useful when a situation involves different scenarios or rules over its set of inputs. For example, in our exercise, the piecewise function is used to describe the velocity of an object over several distinct time intervals.

The usefulness of piecewise functions stems from their ability to accurately model real-world scenarios that have different behaviors at different times or conditions. In our specific case:

• The velocity changes from a constant positive rate, to a decreasing rate, and finally to a constant negative rate after a certain point.
• This variability allows us to address each phase independently, providing a clear understanding of the object's behavior during each time period.

Piecewise functions empower us to tackle complex behaviors by breaking them into manageable chunks. This simplification is crucial when calculating things like position functions, as it involves integrating each velocity "piece" over its respective interval. By mastering the construction and interpretation of piecewise functions, you can gain valuable insights into variable conditions that occur in real-world phenomena.

Integral of Velocity

The **integral of velocity** function gives the position function in calculus. Essentially, integrating the velocity function over time provides insight into how much total distance the object has covered as a function of time. Integrating velocity functions piecewise allows you to find the position for each interval separately, capturing the complete motion of the object.

In solving the position in our exercise:

• For the first interval \(0 \leq t \leq 100\), integrating a constant velocity of 5 gives \(x(t) = 5t\).
• For the second interval \(100 < t \leq 700\), the linear velocity \(v(t) = 6 - \frac{t}{100}\) is integrated, resulting in a quadratic position formula once we add the initial condition.
• In the third interval \(t > 700\), the constant velocity of -1 is integrated, reflecting the object's return path towards the origin.

Each interval’s integration considers the cumulative distance traveled, steering clear of missing out on transitions between different velocity phases. This method ensures a complete and accurate reflection of the object's journey. Understanding integration in this context is pivotal for predicting future positions or verifying when the object might return to its starting point.