Looking at your data, does the general definition of average v= 4t hold up over a small time interval? Explain your reasons for your answers.

The formula \( v = 4t \) seems to represent the relationship between velocity \( v \) and time \( t \). To assess whether this formula holds over small time intervals, we need to evaluate its consistency with the concept of average velocity, particularly when time intervals are small.

### Key Points to Analyze:

1. **Velocity Function**: 
   The equation \( v = 4t \) implies that the velocity is linearly dependent on time. It means that the velocity increases by 4 units for every unit of time. This represents instantaneous velocity at any time \( t \).

2. **Average Velocity**:
   The **average velocity** over a time interval is defined as:
   \[
   \text{Average velocity} = \frac{\Delta x}{\Delta t}
   \]
   where \( \Delta x \) is the change in position (displacement) and \( \Delta t \) is the change in time over the interval. 

   For a time-varying velocity \( v(t) \), we find the displacement by integrating the velocity function:
   \[
   x(t) = \int v(t) \, dt
   \]
   In this case, \( v(t) = 4t \), so the displacement \( x(t) \) is given by:
   \[
   x(t) = \int 4t \, dt = 2t^2 + C
   \]
   Thus, the displacement is proportional to \( t^2 \).

3. **Small Time Intervals**:
   Over a small time interval \( [t, t + \Delta t] \), we can compute the displacement:
   \[
   \Delta x = x(t + \Delta t) - x(t)
   \]
   Substituting the displacement function \( x(t) = 2t^2 \):
   \[
   \Delta x = 2(t + \Delta t)^2 - 2t^2 = 2(t^2 + 2t\Delta t + (\Delta t)^2) - 2t^2 = 4t\Delta t + 2(\Delta t)^2
   \]
   The average velocity is then:
   \[
   \text{Average velocity} = \frac{4t\Delta t + 2(\Delta t)^2}{\Delta t} = 4t + 2\Delta t
   \]
   As \( \Delta t \to 0 \), the average velocity approaches \( 4t \), which is precisely the instantaneous velocity \( v = 4t \).

### Conclusion:
Yes, the general definition of average velocity \( v = 4t \) holds over a small time interval because, as the time interval \( \Delta t \) becomes smaller, the average velocity approaches the instantaneous velocity given by \( v = 4t \). This behavior is consistent with the concept of instantaneous velocity as the limit of average velocity over infinitesimally small time intervals.

Would you like a more detailed explanation or have any follow-up questions?

Here are 5 related questions to consider:
1. How does the relationship between displacement and velocity change if velocity is not a linear function of time?
2. What would the average velocity be over a finite time interval, say from \( t = 1 \) to \( t = 3 \), for this equation?
3. How does the concept of instantaneous velocity differ from average velocity for non-linear velocity functions?
4. What is the geometric interpretation of velocity as a function of time on a velocity-time graph?
5. How would the calculation change if acceleration were present (i.e., if velocity were quadratic in time)?

**Tip**: When dealing with small intervals, always check the limit behavior to see if average quantities approach their instantaneous counterparts.

Looking at your data, does the general definition of average v = 4t hold up over a small time interval? Explain your reasons for your answers.

Explore whether the formula v = 4t for velocity holds true over small time intervals. This detailed analysis covers calculus concepts, including limits, kinematics, and the fundamental theorem of calculus, suitable for students in Grades 11-12.

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