Determine what is happening at each interval in the graph:A-B___________________________________________B-C___________________________________________C-D__________________________________________ Classroom .ll LTE6:14 PM173%A goformative.comshow questionsime Graph MasteryB)40302010o A)D-101510Time (s)5Use for questions #2-#72520Position (m)

Answered: B 40 30 20 10 (A) NO D -10 5 10 15 20…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Question

Determine what is happening at each interval in the graph:

A-B___________________________________________

B-C___________________________________________

C-D__________________________________________

Classroom .ll LTE
6:14 PM
173%
A goformative.com
show questions
ime Graph Mastery
B)
40
30
20
10
o A)
D
-10
15
10
Time (s)
5
Use for questions #2-#7
25
20
Position (m)

Expert Solution

Step 1 :Introduction

Velocity of an object is defined as the change in position of the object to the time period , that is , $v = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$ , where $x_{f}$ is the final position , $x_{i}$ is the initial position , $t_{f}$ is the final time and $t_{i}$ is the initial time.

As velocity is a vector quantity , the direction of motion is also considered , thus the velocity will be negative when an object comes back , that is initial position value is greater than final position value.

If the slope value in a position-time graph gives the velocity of an object .For a linear increasing or decreasing graph the velocity is constant , while if the position does not change the velocity will be zero.

Step 2 : Explanation

From A to B :

The position of the object here changes thus the object is under motion and as the position change is linear and upward ,thus the velocity of the object will be constant and positive.

Calculate the constant velocity by using the formula $v_{A B} = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$ ,where $x_{f} = 40 m$ is the final position , $x_{i} = 0 m$ is the initial position , $t_{f} = 5 s$ is the final time and $t_{i} = 0 s$ is the initial time.

$\begin{array}{rcl} v_{A B} & = & \frac{x_{f} - x_{i}}{t_{f} - t_{i}} \\ = & \frac{40 m - 0 m}{5 s - 0 s} \\ = & 8 m / s \end{array}$

From B to C :

The position of the object here does not change thus the object is stationary and so the velocity of the object will be zero.

Calculate the velocity by using the formula $v_{B C} = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$ ,where $x_{f} = 40 m$ is the final position , $x_{i} = 40 m$ is the initial position , $t_{f} = 15 s$ is the final time and $t_{i} = 5 s$ is the initial time.

$\begin{array}{rcl} v_{B C} & = & \frac{x_{f} - x_{i}}{t_{f} - t_{i}} \\ = & \frac{40 m - 40 m}{15 s - 5 s} \\ = & 0 m / s \end{array}$

From C to D :

The position of the object here changes thus the object is under motion and as the position change is linear and downward ,thus the velocity of the object will be constant and negative.

Calculate the negative velocity by using the formula $v_{C D} = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$ ,where $x_{f} = - 10 m$ is the final position , $x_{i} = 40 m$ is the initial position , $t_{f} = 20 s$ is the final time and $t_{i} = 15 s$ is the initial time.

$\begin{array}{rcl} v_{C D} & = & \frac{x_{f} - x_{i}}{t_{f} - t_{i}} \\ = & \frac{- 10 m - 40 m}{20 s - 15 s} \\ = & - 10 m / s \end{array}$

Step by step

Solved in 3 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions