Math / Calculus

QuestionCompleted: 9 of 11 My score: $7.63 / 11$ pts (69.32\%)
Given the graph of the positive velocity of an object moving along a line, what is the geometric representation of its displacement over a time interval [a, b]?
Let the horizontal axis measure time in terms of t . Choose the correct answer below. A. The object's displacement can be represented geometrically as the area between the graph and the t -axis from $\mathrm{t}=\mathrm{a}$ to $\mathrm{t}=\mathrm{b}$ . B. The object's displacement can be represented geometrically as the difference between the value of the graph at $t=a$ and the value of the graph at $t=b$ . C. The object's displacement can be represented geometrically as the average value of the graph between $t=a$ and $t=b$ . D. The object's displacement can be represented geometrically as the slope of the graph between $t=a$ and $t=b$ .

Studdy Solution

STEP 1

What is this asking? If we know how fast something is moving at different times, how can we figure out how far it's traveled overall? Watch out! Velocity is how fast and in which direction something's moving, while displacement is the overall change in position.
Don't mix them up!

STEP 2

1. Understand Velocity and Displacement
2. Connect Velocity to Area

STEP 3

Imagine a car moving at a constant velocity of $50$ meters per second.
After $2$ seconds, it will have traveled $50 \cdot 2 = 100$ meters.
This calculation is simply the velocity multiplied by the time.

STEP 4

Graphically, this is a rectangle!
The width is the time interval ( $2$ seconds), the height is the velocity ( $50$ meters/second), and the area is the displacement ( $100$ meters).

STEP 5

Now, what if the velocity isn't constant?
We can break the time interval into tiny pieces, where the velocity is almost constant in each tiny piece.

STEP 6

For each tiny piece, the displacement is approximately the velocity at that moment times the tiny time interval.
This is the area of a tiny rectangle!

STEP 7

Adding up the areas of all these tiny rectangles gives us an approximation of the total displacement.
As the rectangles get smaller and smaller, this approximation gets closer and closer to the exact displacement, which is the area under the velocity graph.

STEP 8

The problem gives us a graph of the velocity of an object.
We want to find the object's displacement over the time interval $[a, b]$ .

STEP 9

From our earlier thinking, we know that displacement is represented by the area under the velocity graph.

STEP 10

Specifically, the displacement over the interval $[a, b]$ is the area between the velocity graph and the $t$ -axis from $t = a$ to $t = b$ .

STEP 11

The correct answer is A.
The object's displacement can be represented geometrically as the area between the graph and the $t$ -axis from $t = a$ to $t = b$ .

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Studdy Solution

STEP 1

What is this asking? If we know how fast something is moving at different times, how can we figure out how far it's traveled overall? Watch out! Velocity is how fast and in which direction something's moving, while displacement is the overall change in position.
Don't mix them up!

STEP 2

1. Understand Velocity and Displacement
2. Connect Velocity to Area

STEP 3

Imagine a car moving at a constant velocity of $50$ meters per second.
After $2$ seconds, it will have traveled $50 \cdot 2 = 100$ meters.
This calculation is simply the velocity multiplied by the time.

STEP 4

Graphically, this is a rectangle!
The width is the time interval ( $2$ seconds), the height is the velocity ( $50$ meters/second), and the area is the displacement ( $100$ meters).

STEP 5

Now, what if the velocity isn't constant?
We can break the time interval into tiny pieces, where the velocity is almost constant in each tiny piece.

STEP 6

For each tiny piece, the displacement is approximately the velocity at that moment times the tiny time interval.
This is the area of a tiny rectangle!

STEP 7

Adding up the areas of all these tiny rectangles gives us an approximation of the total displacement.
As the rectangles get smaller and smaller, this approximation gets closer and closer to the exact displacement, which is the area under the velocity graph.

STEP 8

The problem gives us a graph of the velocity of an object.
We want to find the object's displacement over the time interval $[a, b]$ .

STEP 9

From our earlier thinking, we know that displacement is represented by the area under the velocity graph.

STEP 10

Specifically, the displacement over the interval $[a, b]$ is the area between the velocity graph and the $t$ -axis from $t = a$ to $t = b$ .

STEP 11

The correct answer is A.
The object's displacement can be represented geometrically as the area between the graph and the $t$ -axis from $t = a$ to $t = b$ .

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Studdy Solution

STEP 1

What is this asking? If we know how fast something is moving at different times, how can we figure out how far it's traveled overall? Watch out! Velocity is *how fast* and *in which direction* something's moving, while displacement is the *overall change in position*.Don't mix them up!

STEP 2

1. Understand Velocity and Displacement2. Connect Velocity to Area

STEP 3

Imagine a car moving at a *constant* velocity of 505050 meters per second.After 222 seconds, it will have traveled 50⋅2=10050 \cdot 2 = 10050⋅2=100 meters.This calculation is simply the **velocity** multiplied by the **time**.

STEP 4

Graphically, this is a rectangle!The width is the **time interval** (222 seconds), the height is the **velocity** (505050 meters/second), and the *area* is the **displacement** (100100100 meters).

STEP 5

Now, what if the velocity isn't constant?We can break the time interval into tiny pieces, where the velocity is *almost* constant in each tiny piece.

STEP 6

For each tiny piece, the displacement is *approximately* the velocity at that moment times the tiny time interval.This is the area of a tiny rectangle!

STEP 7

Adding up the areas of all these tiny rectangles gives us an *approximation* of the total displacement.As the rectangles get smaller and smaller, this approximation gets closer and closer to the *exact* displacement, which is the area under the velocity graph.

STEP 8

The problem gives us a graph of the velocity of an object.We want to find the object's displacement over the time interval [a,b][a, b][a,b].

STEP 9

From our earlier thinking, we know that displacement is represented by the area under the velocity graph.

STEP 10

Specifically, the displacement over the interval [a,b][a, b][a,b] is the area between the velocity graph and the ttt-axis from t=at = at=a to t=bt = bt=b.

STEP 11

The correct answer is A.The object's displacement can be represented geometrically as the area between the graph and the ttt-axis from t=at = at=a to t=bt = bt=b.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? If we know how fast something is moving at different times, how can we figure out how far it's traveled overall? Watch out! Velocity is how fast and in which direction something's moving, while displacement is the overall change in position.
Don't mix them up!

1. Understand Velocity and Displacement
2. Connect Velocity to Area

Imagine a car moving at a constant velocity of $50$ meters per second.
After $2$ seconds, it will have traveled $50 \cdot 2 = 100$ meters.
This calculation is simply the velocity multiplied by the time.

Graphically, this is a rectangle!
The width is the time interval ( $2$ seconds), the height is the velocity ( $50$ meters/second), and the area is the displacement ( $100$ meters).

Now, what if the velocity isn't constant?
We can break the time interval into tiny pieces, where the velocity is almost constant in each tiny piece.

For each tiny piece, the displacement is approximately the velocity at that moment times the tiny time interval.
This is the area of a tiny rectangle!

Adding up the areas of all these tiny rectangles gives us an approximation of the total displacement.
As the rectangles get smaller and smaller, this approximation gets closer and closer to the exact displacement, which is the area under the velocity graph.

The problem gives us a graph of the velocity of an object.
We want to find the object's displacement over the time interval $[a, b]$ .

Specifically, the displacement over the interval $[a, b]$ is the area between the velocity graph and the $t$ -axis from $t = a$ to $t = b$ .

The correct answer is A.
The object's displacement can be represented geometrically as the area between the graph and the $t$ -axis from $t = a$ to $t = b$ .