The graph shows the velocity v(t) of an object. V(t) 4 3 2 1 5 V(t) 0 1 2 3 4 7 8 9 10 t Calculate the object's displacement between [1, 6]. O 12 O 12.5 O 13 O 13.5 6
The graph shows the velocity v(t) of an object. V(t) 4 3 2 1 5 V(t) 0 1 2 3 4 7 8 9 10 t Calculate the object's displacement between [1, 6]. O 12 O 12.5 O 13 O 13.5 6
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please explain how to solve, thank you!

In the graph above:
- The x-axis represents time t.
- The y-axis represents the velocity v(t).
- The graph spans from t = 1 to t = 10.
The graph shows the following key points:
- From t = 1 to t = 2, the velocity increases linearly from 1 to 2 units.
- From t = 2 to t = 3, the velocity remains constant at 2 units.
- From t = 3 to t = 4, the velocity increases linearly from 2 to 3 units.
- From t = 4 to t = 6, the velocity remains constant at 3 units.
- From t = 6 to t = 7, the velocity decreases linearly from 3 to 2 units.
- From t = 7 to t = 8, the velocity remains constant at 2 units.
- From t = 8 to t = 9, the velocity decreases linearly from 2 to 1 unit.
- From t = 9 to t = 10, the velocity remains constant at 1 unit.
#### Calculation of Displacement
To calculate the displacement of the object between t = 1 and t = 6, we need to find the area under the velocity-time graph in this time interval.
**Area Calculation:**
- Area from t = 1 to t = 2 (Trapezoid): \( \frac{1}{2}(1+2) \times 1 = 1.5 \)
- Area from t = 2 to t = 3 (Rectangle): \( 2 \times 1 = 2 \)
- Area from t = 3 to t = 4 (Trapezoid): \( \frac{1}{2}(2+3) \times 1 = 2.5 \)
- Area from t = 4 to t = 6 (Rectangle): \( 3 \times 2 = 6 \)
**Total Displacement:**
Total displacement between t = 1 and t = 6 = \(1.5 + 2 + 2.5 + 6 = 12\) units
**Multiple Choice Answer:**
- \(\bigcirc\) 12
- \( \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F883933e7-6864-477d-8bb2-712659e12b95%2Febb9e136-3dc0-49f9-81d5-ef385e9fe9a2%2Fw1dzzgp_processed.png&w=3840&q=75)
Transcribed Image Text:### Velocity and Displacement Calculation from Graph
#### The graph shows the velocity v(t) of an object.

In the graph above:
- The x-axis represents time t.
- The y-axis represents the velocity v(t).
- The graph spans from t = 1 to t = 10.
The graph shows the following key points:
- From t = 1 to t = 2, the velocity increases linearly from 1 to 2 units.
- From t = 2 to t = 3, the velocity remains constant at 2 units.
- From t = 3 to t = 4, the velocity increases linearly from 2 to 3 units.
- From t = 4 to t = 6, the velocity remains constant at 3 units.
- From t = 6 to t = 7, the velocity decreases linearly from 3 to 2 units.
- From t = 7 to t = 8, the velocity remains constant at 2 units.
- From t = 8 to t = 9, the velocity decreases linearly from 2 to 1 unit.
- From t = 9 to t = 10, the velocity remains constant at 1 unit.
#### Calculation of Displacement
To calculate the displacement of the object between t = 1 and t = 6, we need to find the area under the velocity-time graph in this time interval.
**Area Calculation:**
- Area from t = 1 to t = 2 (Trapezoid): \( \frac{1}{2}(1+2) \times 1 = 1.5 \)
- Area from t = 2 to t = 3 (Rectangle): \( 2 \times 1 = 2 \)
- Area from t = 3 to t = 4 (Trapezoid): \( \frac{1}{2}(2+3) \times 1 = 2.5 \)
- Area from t = 4 to t = 6 (Rectangle): \( 3 \times 2 = 6 \)
**Total Displacement:**
Total displacement between t = 1 and t = 6 = \(1.5 + 2 + 2.5 + 6 = 12\) units
**Multiple Choice Answer:**
- \(\bigcirc\) 12
- \( \
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