Question

94. [Use of Tech] Skydiving A skydiver has a downward velocity given by v(t) = V_T [(1 - e^(-2gt/V_T))/(1 + e^(-2gt/V_T))],
where t = 0 is the instant the skydiver starts falling, g = 9.8 m/s² is the acceleration due to gravity, and V_T is the terminal velocity of the skydiver.
c. Verify by integration that the position function is given by
s(t) = V_T t + (V_T²/g) ln[(1 + e^(-2gt/V_T))/2],
where s'(t) = v(t) and s(0) = 0.

Accepted Answer

Start with the given velocity function:

$$v(t) = V_T \left( \frac{1 - e$$^{-\frac{2gt}$${V_T}}}{1 + e$$^{-\frac{2gt}$${V_T}}} ight).

$$Since velocity is the derivative of position, we have $$s'(t) = v(t)$$, so to find $$s(t)$$, integrate $$v(t)$$ with respect to $$t. $$Rewrite the velocity function to simplify the integral. Notice that

$$\frac{1 - e$$^{-x}$$}{1 + e$$^{-x}$$} = 	anh\left( \frac{x}{2} ight)$$,

where $$x = \frac{2gt}{V_T}. So$$,

$$v(t) = V_T 	anh\left( \frac{gt}{V_T} ight).

$$This substitution makes the integral more straightforward. Set up the integral for position:

$$s(t) = \int v(t) \, dt = \int V_T 	anh\left( \frac{gt}{V_T} ight) dt.

$$Since $$V_T is $$constant, factor it out:

$$s(t) = V_T \int 	anh\left( \frac{gt}{V_T} ight) dt. $$Use substitution to integrate:

Let $$u = \frac{gt}{V_T}$$, so $$du = \frac{g}{V_T} dt or$$ $$dt = \frac{V_T}{g} du.

$$Rewrite the integral:

$$s(t) = V_T \int 	anh(u) \cdot \frac{V_T}{g} du = \frac{V_T^2$}{g} \int 	anh(u) du. $$Recall the integral of hyperbolic tangent:

$$\int 	anh(u) du = \ln(\cosh(u)) + C.

$$Therefore,

$$s(t) = \frac{V_T^2$}{g} \ln(\cosh(u)) + C = \frac{V_T^2$}{g} \ln\left( \cosh\left( \frac{gt}{V_T} ight) ight) + C.

$$Use the initial condition $$s(0) = 0 to $$solve for $$C$$ and express $$\cosh in $$terms of exponentials to match the given formula.

Verified video answer for a similar problem:

Key Concepts

Relationship Between Velocity and Position via Integration

Integration of Exponential Functions

Initial Conditions and Determining Constants of Integration