Question

Suppose an object moves along a line at \(15 \mathrm{m} / \mathrm{s},\) for \(0 \leq t<2\) and at \(25 \mathrm{m} / \mathrm{s},\) for \(2 \leq t \leq 5,\) where \(t\) is measured in seconds. Sketch the graph of the velocity function and find the displacement of the object for \(0 \leq t \leq 5\).

Short answer

Based on the given velocity function, which is a piecewise function with \(15 \mathrm{m/s}\) from \(0 \leq t < 2\) and \(25 \mathrm{m/s}\) from \(2 \leq t \leq 5\), we can sketch the graph as two horizontal lines, one at \(15 \mathrm{m/s}\) and another at \(25 \mathrm{m/s}\). After integrating the function over each time interval and adding the results, we find that the total displacement of the object during the entire time interval is 105 meters.

Step-by-step solution

Set up the graph and the velocity function

To sketch the graph, we should represent the velocity function as a piecewise function: $$ v(t) = \begin{cases} 15 \ \mathrm{m/s} & \text{for} \ 0 \leq t < 2 \\ 25 \ \mathrm{m/s} & \text{for} \ 2 \leq t \leq 5 \end{cases} $$ This will be two horizontal lines, one at \(15 \mathrm{m/s}\) and another at \(25 \mathrm{m/s}\).

Sketch the graph of the velocity function

Plot the piecewise function on the graph, where the x-axis represents the time in seconds and the y-axis represents the velocity in meters per second. Draw a horizontal line at \(15 \mathrm{m/s}\) for the interval \(0 \leq t < 2\), and another horizontal line at \(25 \mathrm{m/s}\) for the interval \(2 \leq t \leq 5\). The graph is a simple representation of the object's velocity for each time interval.

Find the displacement in each time interval

To find the displacement, we need to integrate the velocity function over the interval \([0, 5]\). Since the function is piecewise, we will integrate over each piece and then add the results. For the first interval, \(0 \leq t < 2\): $$ \int_{0}^{2} 15 \, dt = 15 \int_{0}^{2} dt = 15 (t \Big|_{0}^{2}) = 15(2) = 30 \, \mathrm{m} $$ For the second interval, \(2 \leq t \leq 5\): $$ \int_{2}^{5} 25 \, dt = 25 \int_{2}^{5} dt = 25 (t \Big|_{2}^{5}) = 25(5 - 2) = 75 \, \mathrm{m} $$

Calculate the total displacement

To find the total displacement, we need to add the displacement found in each time interval: Total displacement = displacement in the first interval + displacement in the second interval $$ = 30 \, \mathrm{m} + 75 \, \mathrm{m} = 105 \, \mathrm{m} $$ So, the total displacement of the object during the entire time interval for \(0 \leq t \leq 5\) is 105 meters.

Key concepts

Piecewise Velocity Function

When an object's velocity varies at different time intervals, we can use a piecewise velocity function to describe its motion. In this way, the piecewise function captures discrete changes in velocity at specified times, allowing for a segmented yet accurate description of the object’s movement. For example, if an object moves at one velocity for a certain period, and then abruptly changes to a different one for the next period, this is effectively represented by a piecewise function. The examination of such an object moving at 15 m/s for the first 2 seconds, then shifting to 25 m/s up to the 5-second mark, produces two distinct velocities that are each applicable to their specific time frames.

In essence, each segment of our function corresponds to a distinct, uniform velocity over an interval, with our example providing two such intervals: one from 0 to just before 2 seconds, and another from 2 to 5 seconds. Understanding piecewise functions is valuable, as they mimic real-life situations where velocities change due to factors like traffic lights or speed zones.

Graphing Velocity

Graphing velocity is a visual means to represent an object’s speed over time. A velocity-time graph is particularly straightforward when depicting piecewise functions, as the velocity remains constant within given intervals, creating horizontal line segments on the graph. These line segments are drawn parallel to the time-axis at heights that correspond to the constant velocities. In the context of our exercise, we picture this as one line at 15 m/s from 0 to 2 seconds, then at a distinct level of 25 m/s from 2 to 5 seconds. This type of graph allows us to grasp the rates of change at a glance; it shows when and how abrupt changes in speed occur.

When interpreting these graphs, the steepness of the line doesn’t vary within these constant periods, since the velocity doesn’t change. The sudden leaps from one velocity to another are not reflected with diagonal lines, as there are no intermediate speeds—it’s a direct jump from one speed to another. This visualization bridges the gap between numerical data and comprehensible patterns, simplifying the complexity of an object's motion.

Integration to Find Displacement

To transition from velocity to displacement, we employ integration, a fundamental concept in calculus. By integrating a velocity function over a given time interval, we can determine the total displacement—or the overall distance—covered by an object. Integration captures the accumulated change, effectively summing minute displacements over time to reach a holistic distance value.

In our example, the integration over each constant-velocity period yields simple calculations: the area under each horizontal line segment on the graph. The first integration from 0 to 2 seconds takes the constant velocity 15 m/s and multiplies it by the width of the time interval, giving a displacement of 30 meters. Similarly, the second integration from 2 to 5 seconds uses the constant velocity of 25 m/s multiplied by the width of 3 seconds, resulting in 75 meters. Adding these displacements together reveals that the object traveled a total of 105 meters during the 5-second time span.

This process of integration isn’t just a mechanical operation; it is the key to unlocking the 'story' of an object’s journey from its velocity profile.