https://file.aiquickdraw.com/mathbot/file/1727188630568sdw071cq.jpg

The graph shows the velocity (\(v\)) of Hans as a function of time (\(t\)) during his cycling trip. Based on this, we can calculate the acceleration and the distance traveled. Let's address the two questions:

### a) How large is the acceleration during the first 12 seconds?

Acceleration is the change in velocity divided by the time taken. The slope of the velocity-time graph during the first 12 seconds gives us the acceleration.

From the graph:
- At \( t = 0 \) seconds, \( v = 0 \, \text{m/s} \)
- At \( t = 12 \) seconds, \( v = 8 \, \text{m/s} \)

The formula for acceleration (\( a \)) is:
\[
a = \frac{\Delta v}{\Delta t}
\]

Substituting the values:
\[
a = \frac{8 \, \text{m/s} - 0 \, \text{m/s}}{12 \, \text{s}} = \frac{8}{12} = 0.67 \, \text{m/s}^2
\]

Thus, the acceleration during the first 12 seconds is \( 0.67 \, \text{m/s}^2 \).

### b) How far does Hans travel to his friend?

The distance traveled is the area under the velocity-time graph. We can break this graph into three regions:

1. **Region 1 (Acceleration phase)**: From 0 to 12 seconds, the graph forms a right triangle with base 12 seconds and height 8 m/s. The area of this triangle is:
   \[
   A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \, \text{s} \times 8 \, \text{m/s} = 48 \, \text{m}
   \]

2. **Region 2 (Constant velocity phase)**: From 12 to 24 seconds, the velocity is constant at 8 m/s, forming a rectangle. The area of this rectangle is:
   \[
   A_2 = \text{base} \times \text{height} = 12 \, \text{s} \times 8 \, \text{m/s} = 96 \, \text{m}
   \]

3. **Region 3 (Deceleration phase)**: From 24 to 30 seconds, the graph forms another right triangle with base 6 seconds and height 8 m/s. The area of this triangle is:
   \[
   A_3 = \frac{1}{2} \times 6 \, \text{s} \times 8 \, \text{m/s} = 24 \, \text{m}
   \]

The total distance traveled is the sum of these areas:
\[
\text{Total distance} = A_1 + A_2 + A_3 = 48 \, \text{m} + 96 \, \text{m} + 24 \, \text{m} = 168 \, \text{m}
\]

Thus, Hans traveled a total of **168 meters** to reach his friend.

---

Do you want more details or have any questions? Here are some related questions to expand your understanding:

1. What would happen to the acceleration if Hans reached 10 m/s in 12 seconds instead of 8 m/s?
2. How would you calculate the deceleration during the final phase of the trip?
3. What if Hans maintained his velocity at 8 m/s for 15 seconds instead of 12 seconds? How would the total distance change?
4. How would the graph change if the maximum velocity Hans reached was 12 m/s?
5. What if Hans's velocity-time graph showed a non-linear acceleration? How would you calculate the distance?

**Tip:** The area under a velocity-time graph always represents the distance traveled.

Hans cyklar till sin kompis. Grafen visar hastigheten som funktion av tiden under cykelturen. a) Hur stor är accelerationen under de 12 första sekunderna? b) Hur lång väg har han till sin kompis?

Learn how to calculate acceleration and the distance traveled using a velocity-time graph. This problem demonstrates key kinematic concepts such as acceleration and total distance covered through graphical analysis, suitable for students in Grades 9-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation