Worksheet-Velocity-Graph - Section:______ Name: Date:______ - Studocu - Free Printable

Problem Analysis and Solution

The provided image contains a velocity-time graph of a moving car. We are tasked with analyzing the graph to answer several questions. Let's solve each part step by step.

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#### Part I: Velocity-Time Graph Analysis

The graph shows the velocity of a car as a function of time, with:
- Velocity (y-axis) in kilometers per hour (km/h).
- Time (x-axis) in hours.

#### 1. Questions Based on the Graph

##### a. At what time was the car stopped?
- The car is stopped when its velocity is 0 km/h.
- From the graph, the velocity is 0 km/h at time = 0.7 hours.

Answer: \( \boxed{0.7 \text{ hours}} \)

##### b. At what time did the car have the greatest velocity?
- The greatest velocity occurs at the highest point on the graph.
- From the graph, the highest velocity is reached at time = 0.2 hours.

Answer: \( \boxed{0.2 \text{ hours}} \)

##### c. What was the greatest velocity?
- The greatest velocity corresponds to the y-coordinate of the highest point on the graph.
- From the graph, the greatest velocity is 60 km/h.

Answer: \( \boxed{60 \text{ km/h}} \)

##### d. At what time(s) was the car accelerating?
- Acceleration occurs when the velocity is changing (i.e., the slope of the velocity-time graph is non-zero).
- From the graph:
- The car is accelerating from time = 0 to 0.2 hours (velocity increasing).
- The car is also accelerating from time = 0.8 to 1.0 hours (velocity increasing).

Answer: \( \boxed{0 \text{ to } 0.2 \text{ hours and } 0.8 \text{ to } 1.0 \text{ hours}} \)

##### e. How fast was the car going at 1.0 hours?
- At time = 1.0 hours, read the velocity from the graph.
- From the graph, the velocity at 1.0 hours is 40 km/h.

Answer: \( \boxed{40 \text{ km/h}} \)

##### f. What is the acceleration at 0.9 hours?
- Acceleration is the slope of the velocity-time graph.
- At time = 0.9 hours, the graph shows a straight line with a positive slope.
- To calculate the slope:
- Choose two points on the line near \( t = 0.9 \):
- Point 1: \( (0.8, 0) \)
- Point 2: \( (1.0, 40) \)
- Slope (acceleration) = \( \frac{\Delta v}{\Delta t} = \frac{40 - 0}{1.0 - 0.8} = \frac{40}{0.2} = 200 \text{ km/h}^2 \).

Answer: \( \boxed{200 \text{ km/h}^2} \)

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#### Part II: Graphing Positions and Computing Velocities

This part requires data from a table that is not provided in the image. Since the table is missing, we cannot proceed with this part. If you can provide the table, I can help you complete this section as well.

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Final Answers for Part I:

1. a. \( \boxed{0.7 \text{ hours}} \)
b. \( \boxed{0.2 \text{ hours}} \)
c. \( \boxed{60 \text{ km/h}} \)
d. \( \boxed{0 \text{ to } 0.2 \text{ hours and } 0.8 \text{ to } 1.0 \text{ hours}} \)
e. \( \boxed{40 \text{ km/h}} \)
f. \( \boxed{200 \text{ km/h}^2} \)

If you have any further questions or need clarification, feel free to ask!