v(t) = 190(1 − e^−0.24t) Find the velocity after 4 s and after 14 s

Given the velocity function \( v(t) = 190(1 - e^{-0.24t}) \), we are tasked with finding the velocity after 4 seconds and after 14 seconds.

### Step-by-Step Calculation:

1. **For \( t = 4 \) seconds**:

\[
v(4) = 190(1 - e^{-0.24 \times 4})
\]
\[
v(4) = 190(1 - e^{-0.96})
\]
Now, let's calculate this value.

2. **For \( t = 14 \) seconds**:

\[
v(14) = 190(1 - e^{-0.24 \times 14})
\]
\[
v(14) = 190(1 - e^{-3.36})
\]

Now, I'll compute the actual values for \( v(4) \) and \( v(14) \).The velocity after 4 seconds is approximately \( 117.25 \, \text{m/s} \), and the velocity after 14 seconds is approximately \( 183.40 \, \text{m/s} \).

Would you like further details or explanations on these results?

Here are 5 related questions to expand your understanding:
1. How does the rate constant \( 0.24 \) affect the velocity over time?
2. What is the velocity as \( t \) approaches infinity?
3. How would you compute the acceleration from this velocity function?
4. What happens to the velocity function if you change the initial constant \( 190 \)?
5. How can you graph this velocity function over time?

**Tip:** When dealing with exponential functions, the term \( e^{-kt} \) approaches zero as \( t \) increases, leading the velocity to approach a constant value.

Explore how to solve for velocity using the exponential decay function v(t) = 190(1 − e^−0.24t) at 4 seconds and 14 seconds. This solution involves calculating the velocity values by applying exponential formulas. Suitable for high school students in Grades 10-12.

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