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Suppose that the velocity [tex]v(t)[/tex] (in [tex]m/s[/tex]) of a sky diver falling near the Earth's surface is given by the following function, where time [tex]t[/tex] is measured in seconds:

[tex]v(t) = 56\left(1 - e^{-0.24t}\right)[/tex]

Find the initial velocity of the sky diver and the velocity after 5 seconds. Round your answers to the nearest whole number as necessary.

- Initial velocity: [tex]\square[/tex] [tex]m/s[/tex]
- Velocity after 5 seconds: [tex]\square[/tex] [tex]m/s[/tex]


Sagot :

To solve the problem of finding the initial velocity of the skydiver and the velocity after 5 seconds, we'll analyze the given velocity function:

[tex]\[ v(t) = 56 \left(1 - e^{-0.24 t} \right) \][/tex]

### Step-by-Step Solution:

1. Initial Velocity:

The initial velocity is the velocity at [tex]\( t = 0 \)[/tex] seconds. We will substitute [tex]\( t = 0 \)[/tex] into the velocity function.

[tex]\[ v(0) = 56 \left(1 - e^{-0.24 \cdot 0}\right) \][/tex]

Simplifying inside the parentheses:

[tex]\[ e^{-0.24 \cdot 0} = e^0 = 1 \][/tex]

Therefore,

[tex]\[ v(0) = 56 \left(1 - 1\right) = 56 \cdot 0 = 0 \][/tex]

So, the initial velocity is:

[tex]\[ \boxed{0} \, \text{m/s} \][/tex]

2. Velocity after 5 Seconds:

Next, we want to find the velocity at [tex]\( t = 5 \)[/tex] seconds. We substitute [tex]\( t = 5 \)[/tex] into the velocity function.

[tex]\[ v(5) = 56 \left(1 - e^{-0.24 \cdot 5}\right) \][/tex]

Simplifying the exponent:

[tex]\[ -0.24 \cdot 5 = -1.2 \][/tex]

Thus,

[tex]\[ v(5) = 56 \left(1 - e^{-1.2} \right) \][/tex]

We use the approximate value for [tex]\( e^{-1.2} \)[/tex] which is around 0.3012.

Therefore,

[tex]\[ v(5) = 56 \left(1 - 0.3012\right) \][/tex]

[tex]\[ v(5) = 56 \cdot 0.6988 \][/tex]

Calculating this product:

[tex]\[ v(5) \approx 39.9296 \][/tex]

Rounding to the nearest whole number gives:

[tex]\[ \boxed{39} \, \text{m/s} \][/tex]

### Summary:

- The initial velocity is [tex]\( \boxed{0} \)[/tex] m/s.
- The velocity after 5 seconds is [tex]\( \boxed{39} \)[/tex] m/s.