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?

Short answer

The object is farthest to the right at position 1150 (occurs at \(t=300\)) and returns to the origin at \(t=1850\).

Step-by-step solution

Understand Velocity Function

The given velocity function changes at different points in time. For \(0 \leq t \leq 100\), the velocity \(v(t) = 5\). For \(100 < t \leq 700\), the velocity is \(v(t) = 6 - \frac{t}{100}\). For \(t > 700\), the velocity is constant at \(-1\). This tells us how the object moves over these intervals.

Find Position for \(0 \leq t \leq 100\)

During the interval \(0 \leq t \leq 100\), the velocity is constant at 5. The position function, \(s(t)\), can be found by integrating the velocity: \(s(t) = \int 5 \, dt = 5t + C\). Since the object starts at the origin, \(C = 0\). So, \(s(t) = 5t\) for \(0 \leq t \leq 100\).

Find Position for \(100 < t \leq 700\)

At \(t = 100\), the object is at position \(s(100) = 500\). For \(100 < t \leq 700\), integrate \(v(t) = 6 - \frac{t}{100}\): \(s(t) = \int (6 - \frac{t}{100}) \, dt = 6t - \frac{t^2}{200} + C\). Solve for \(C\) using \(s(100) = 500\): \(500 = 600 - 50 + C \Rightarrow C = -50\). So, \(s(t) = 6t - \frac{t^2}{200} - 50\) for \(100 < t \leq 700\).

Find Position for \(t > 700\)

At \(t = 700\), the object is at position \(s(700) = 1150\). For \(t > 700\), the velocity is constant at \(-1\), so the position is \(s(t) = \int -1 \, dt = -t + C\). Solve for \(C\) using \(s(700) = 1150\): \(1150 = -700 + C \Rightarrow C = 1850\). So, \(s(t) = -t + 1850\) for \(t > 700\).

Determine the Farthest Right Position

Examine the position function to find the maximum position. For \(100 < t \leq 700\), the position \(s(t) = 6t - \frac{t^2}{200} - 50\) is a quadratic function. It achieves its maximum at the vertex. The vertex formula for \(at^2 + bt + c\) is \(t = -\frac{b}{2a}\). Substituting \(a = -\frac{1}{200}, b = 6\), \(t = \frac{600}{-1} = 300\). Substitute \(t = 300\) into \(s(t)\) to find \(s(300) = 1150\), which occurs at \(t = 300\).

Determine If the Object Returns to Origin

For \(t > 700\), the position function is \(s(t) = -t + 1850\). Solve \(-t + 1850 = 0\) to find \(t = 1850\). Therefore, the object returns to the origin at \(t = 1850\).

Key concepts

Velocity Function

In calculus, the velocity function is crucial to understanding the motion of an object over time. It describes how fast the object moves and in which direction at any given point in time. For the given exercise, the velocity is defined piecewise. This means that it is split into different expressions depending on the time interval:

• From 0 to 100 seconds, the object moves at a constant speed of 5 units per second. This indicates uniform motion.
• From 100 to 700 seconds, the velocity decreases linearly from 6 to -1 units, suggesting the object is slowing down.
• After 700 seconds, the velocity becomes -1, indicating the object moves backward at a constant speed.

Understanding the velocity function allows us to predict not only how the object's speed changes but also when and how its direction of motion alters. The piecewise nature of the velocity function reflects different phases of motion.

Position Function

The position function helps us determine where an object is located at any given time. It is derived from the velocity function through integration. Since the velocity function provides the rate of change of position, integrating it gives us the object's position over time:

• For the interval 0 to 100 seconds with constant velocity, the position is straightforward: \( s(t) = 5t \). This comes from integrating the constant velocity, remembering that initially \( C = 0 \) because the object starts at the origin.
• For the interval 100 to 700 seconds, the velocity function is more complex, so the integration gives \( s(t) = 6t - \frac{t^2}{200} - 50 \). The constant \( C \) is computed using the known position at time \( t = 100 \).
• After 700 seconds, where the velocity is -1, the position becomes \( s(t) = -t + 1850 \), adjusted with the constant calculated at \( t = 700 \).

These formulas allow us to track not just the position of the object at any time, but also to see how its route evolves through different phases.

Integration

Integration is a fundamental concept in calculus used to find areas, volumes, and in this case, to determine the position from velocity. When you integrate a velocity function, you effectively add up all the small changes in position to find the total position at a given time. This is particularly useful when the velocity is variable, as in our exercise:

• For a constant velocity, like 5, the position from integration simply multiplies velocity by time: \( \int 5 \, dt = 5t + C \).
• For a non-constant velocity like \( 6 - \frac{t}{100} \), integration yields a quadratic function: \( \int (6 - \frac{t}{100}) \, dt = 6t - \frac{t^2}{200} + C \).
• For periods with negative velocity, integration similarly finds the cumulative effect of moving backwards.

So, integration turns the rate of change given by velocity into the total change, or position, over time. Each integration constant \( C \) is determined based on initial conditions or known points.

Quadratic Function

A quadratic function is a polynomial function of degree 2 and is usually expressed in the form \( ax^2 + bx + c \). In this exercise, the position function for the interval 100 to 700 seconds is quadratic due to the velocity function's linear form. Here’s why this is significant:

• The formula \( s(t) = 6t - \frac{t^2}{200} - 50 \) contains a term \( \frac{t^2}{200} \), showing a parabolic shape.
• Quadratic functions have a vertex, which is the highest or lowest point, depending on the parabola's direction. Finding the vertex is key to understanding maximum or minimum positions.
• In context, determining the vertex of the quadratic position function tells us when and where the object reaches its farthest point beyond the origin, specifically at \( t = 300 \), yielding the maximum position \( s(300) = 1150 \).

Thus, quadratic functions not only help identify maximum positions but also provide insight into the object’s motion in time.