https://file.aiquickdraw.com/mathbot/file/17294551573595qjyfnxn.jpg

The task in the image involves a falling object, specifically a stone, with the following functional relationship for the distance fallen:

\[
s(t) = 5 \cdot t^2
\]

Here:
- \( s(t) \) is the distance (in meters) the stone has fallen after \( t \) seconds.
- \( t \) is the time (in seconds).

The problem asks you to calculate the velocity \( v(t) \) of the stone at \( t = 2 \) seconds. 

To find the velocity, we need to differentiate the distance function \( s(t) \) with respect to time \( t \), since velocity is the derivative of distance:

\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(5 \cdot t^2) = 10 \cdot t
\]

Now, substituting \( t = 2 \) seconds:

\[
v(2) = 10 \cdot 2 = 20 \, \text{m/s}
\]

So, the velocity of the stone after 2 seconds is **20 m/s**.

Would you like more details or explanations on any step?

Here are 5 additional questions related to this topic:
1. What is the velocity at \( t = 3 \) seconds?
2. How far has the stone fallen after 2 seconds?
3. What is the acceleration of the stone?
4. How would the formula change if the initial velocity was not zero?
5. What is the physical meaning of the constant "5" in the equation \( s(t) = 5 \cdot t^2 \)?

**Tip**: In free-fall problems, the acceleration due to gravity is usually approximated as \( g = 9.81 \, \text{m/s}^2 \), but here the constant "5" simplifies the calculation!

Berechne die Geschwindigkeit in Metern pro Sekunde (m/s), die der Stein nach einer Fallzeit von t = 2 Sekunden hat!

Learn how to calculate the velocity of a falling stone at t = 2 seconds, using the function s(t) = 5 * t^2 for the distance fallen. This problem involves differentiation and basic kinematics, suitable for students in Grades 9-12.

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