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.