Chapter 6: Problem 34
Find the distance traveled by the object on the given interval by finding the areas of the appropriate geometric region. $$ v=f(t)=3,[2,7] $$
Short Answer
Expert verified
The distance traveled is 15 units.
Step by step solution
01
Identify the Problem
We are given a velocity function \(v(t) = 3\) over the interval \([2, 7]\). This means the velocity is constant at 3 units per time unit. To find the distance traveled, we need to determine the area under the velocity-time graph on this interval.
02
Understand the Geometric Shape
Since the velocity is constant, the graph of \(v(t) = 3\) is a horizontal line at \(v = 3\). Between the interval [2,7], this line creates a rectangle with height 3 and width given by the interval range, which is \(7 - 2\).
03
Calculate the Width
Determine the width of the rectangle. The interval \([2, 7]\) gives a width of \(7 - 2 = 5\) units.
04
Calculate the Area
The area of a rectangle is given by \(\text{Area} = \text{Height} \times \text{Width}\). Here, the height is 3, and the width is 5. So, \(\text{Area} = 3 \times 5 = 15\).
05
Relate Area to Distance Traveled
In this context, the area under the velocity-time graph represents the distance traveled by the object. Hence, the distance traveled is 15 units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity-Time Graph
A velocity-time graph provides a visual representation of how an object's velocity changes over time. It's a crucial tool in understanding motion. Here, the **y-axis** represents velocity, while the **x-axis** represents time. Each point on the graph shows the object's velocity at a specific time.
For example, in our problem, the function given is a constant velocity function, denoted as \(v(t) = 3\). This implies that regardless of time, the object's velocity remains steady at 3 units per time unit. A graph of this function would appear as a horizontal line at the height of 3 on the velocity axis. This kind of graph simplifies distance calculations as the area under the curve becomes a straightforward rectangle.
Velocity-time graphs help in identifying:
For example, in our problem, the function given is a constant velocity function, denoted as \(v(t) = 3\). This implies that regardless of time, the object's velocity remains steady at 3 units per time unit. A graph of this function would appear as a horizontal line at the height of 3 on the velocity axis. This kind of graph simplifies distance calculations as the area under the curve becomes a straightforward rectangle.
Velocity-time graphs help in identifying:
- If the velocity is increasing, decreasing, or constant
- The acceleration of the object (constant velocity indicates no acceleration)
- The total displacement or distance covered, by calculating the area under the graph
Area Under the Curve
The area under the curve on a velocity-time graph represents the total distance an object travels over a specific time interval. This is because the area effectively encapsulates the product of velocity and time, which results in distance.
In our scenario, since the velocity \(v(t) = 3\) is constant between the times of 2 and 7, the graph forms a rectangle rather than a complex shape. The area under this constant line from \(t = 2\) to \(t = 7\) directly translates to the total distance covered.
To visualize:
In our scenario, since the velocity \(v(t) = 3\) is constant between the times of 2 and 7, the graph forms a rectangle rather than a complex shape. The area under this constant line from \(t = 2\) to \(t = 7\) directly translates to the total distance covered.
To visualize:
- The base of the rectangle lies along the time axis, extending from 2 to 7 (a width of 5 units).
- The height of the rectangle reaches up to the constant velocity value, which is 3.
Rectangular Area Method
The rectangular area method is a straightforward technique of finding the area under a curve, specifically when dealing with constant functions, as in our exercise. This involves treating the simple geometry beneath the curve as a rectangle.
Here's how the process works with our constant velocity function \(v(t) = 3\):
Here's how the process works with our constant velocity function \(v(t) = 3\):
- **Width of the Rectangle:** The width is determined by the interval on the time axis, hence \(7 - 2 = 5\).
- **Height of the Rectangle:** The graph of \(v(t) = 3\) is a horizontal line at height 3, representing the velocity's constant value.
- **Area Calculation:** The formula for the area of the rectangle is \(\text{Area} = \text{Height} \times \text{Width}\). Substituting the known values gives \(3 \times 5 = 15\).