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.