I am unsure of how to solve this problem. Any help would be greatly appreciated. Thank you. ### Position-Time Graph AnalysisThe graph shown represents the position-time relationship of a toy moving over a period of time.**Position-Time Graph Details:**- **Axes:** - The horizontal axis (x-axis) represents time in seconds (sec), ranging from 0 to 2.00 seconds. - The vertical axis (y-axis) shows position in meters (m), ranging from 0 to 60 meters.- **Graph Description:** - The graph starts at the origin point (0,0), indicating the toy begins its movement from the starting position with a time of 0 seconds. - The toy's position increases rapidly and then begins to plateau as time progresses, indicating a decrease in velocity over time.**Questions:**a. **At what time is the toy going at the greatest velocity?** - The toy is moving at the greatest velocity where the slope of the graph is steepest. This occurs between 0 seconds and approximately 0.75 seconds.b. **How fast is it traveling at that time?** - The greatest velocity can be estimated by calculating the slope of the steepest part of the graph (rise over run). Between 0 and 0.75 seconds, the position changes from 0 to about 30 meters. Therefore, the velocity is approximately: \[ \text{Velocity} = \frac{30 \, \text{m}}{0.75 \, \text{sec}} \approx 40 \, \text{m/s} \]c. **What is the average velocity for the first 1 sec?** - The average velocity is calculated by dividing the total change in position by the total time taken. At 1 second, the position is about 40 meters, so the average velocity is: \[ \text{Average Velocity} = \frac{40 \, \text{m}}{1 \, \text{sec}} = 40 \, \text{m/s} \]This analysis provides an understanding of how to determine velocity from a position-time graph by examining the slope and calculating average velocity over a given time interval.

Name: I am unsure of how to solve this problem. Any help would be greatly ap
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Description: I am unsure of how to solve this problem. Any help would be greatly appreciated. Thank you. ### Position-Time Graph AnalysisThe graph shown represents the position-time relationship of a toy moving over a period of time.**Position-Time Graph Details:

Answered: a. At what time is the toy going at the…

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I am unsure of how to solve this problem. Any help would be greatly appreciated. Thank you.

### Position-Time Graph Analysis

The graph shown represents the position-time relationship of a toy moving over a period of time.

**Position-Time Graph Details:**

- **Axes:**
- The horizontal axis (x-axis) represents time in seconds (sec), ranging from 0 to 2.00 seconds.
- The vertical axis (y-axis) shows position in meters (m), ranging from 0 to 60 meters.

- **Graph Description:**
- The graph starts at the origin point (0,0), indicating the toy begins its movement from the starting position with a time of 0 seconds.
- The toy's position increases rapidly and then begins to plateau as time progresses, indicating a decrease in velocity over time.

**Questions:**

a. **At what time is the toy going at the greatest velocity?**
- The toy is moving at the greatest velocity where the slope of the graph is steepest. This occurs between 0 seconds and approximately 0.75 seconds.

b. **How fast is it traveling at that time?**
- The greatest velocity can be estimated by calculating the slope of the steepest part of the graph (rise over run). Between 0 and 0.75 seconds, the position changes from 0 to about 30 meters. Therefore, the velocity is approximately:
\[
\text{Velocity} = \frac{30 \, \text{m}}{0.75 \, \text{sec}} \approx 40 \, \text{m/s}
\]

c. **What is the average velocity for the first 1 sec?**
- The average velocity is calculated by dividing the total change in position by the total time taken. At 1 second, the position is about 40 meters, so the average velocity is:
\[
\text{Average Velocity} = \frac{40 \, \text{m}}{1 \, \text{sec}} = 40 \, \text{m/s}
\]

This analysis provides an understanding of how to determine velocity from a position-time graph by examining the slope and calculating average velocity over a given time interval.

Expert Solution

Step 1

The slope of position-time curve at any gives the velocity at that point, where the slope is maximum then the velocity is also maximum.

a. From curve it is clear that at time 0.45 s the slope is maximum and so does the velocity.

b. form curve it is seen that the slope is approximately same between time 0.4 s to 0.45 s, therefore the slope of line between these times gives the required velocity.

Formula of slope is :

$s l o p e = \frac{c h a n g e a l o n g y a x i s}{c h a n g e a l o n g x a x i s}$

From curve position at 0.4 s is nearly 15 m and at 0.405 s it is 20 m.

hence using the above formula for the given curve, the velocity will be:

$v = \frac{20 - 15}{0.45 - 0.4} v = 100 m / s$

hence it is moving with nearly 100 m/s.

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