Question

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g bv, where v is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time? A) v = g(1-e^-bt)/b B) v = (ge^bt)/b C) v = (g+a)t/b

Answer 1

Answer:

A) $\frac{g}{b}(1-e^{-bt})$

Explanation:

Since a = g - bv,

We can substitute a = dv/dt into the equation.

Then, the equation will be like dv/dt = g - bv.

So we got first order differential equation.

As known, v = 0 at t = 0, and v = g/b at t = ∞.

Since $\frac{dv}{dt}= g - bv = b( \frac{g}{b} - v)$ ⇒ $\frac{dv}{ \frac{g}{b} - v}= bdt$

So take the integral of both side.

$- ln (\frac{g}{b} - v) = bt + C$

Since for t=0, v = 0 ⇒ $C =- ln (\frac{g}{b})$

$v = \frac{g}{b} + e^{-bt-ln(\frac{g}{b})} = \frac{g}{b}- \frac{g}{b}e^{-bt} = \frac{g}{b}(1-e^{-bt})$

Answer 2

The correct option for the expression of speed as an explicit function of time is option A

A) v = g·(1 - $e^{-b \cdot t$)/b

The reason why option A is correct is given as follows;

Known:

The initial velocity of the object at time t = 0 is v = 0 (object at rest)

The function that represents the acceleration is a = g - b·v

Where;

v = The speed of the object at the given instant

b = A constant term

By considering the limiting case for time t, we have;

At very large values of t, the velocity will increase such that we have;

$\lim \limits_{t \to \infty} a = 0$ therefore,  $\lim \limits_{t \to \infty} g - b\cdot v = 0$ and $\lim \limits_{t \to \infty} \left( v_{max} = \dfrac{g}{b} \right)$

The given equation can be rewritten as follows, to express the equation in terms the velocity;

$a = b \cdot \left(\dfrac{g}{b} - v \right) = b \cdot \left(v_{max} - v \right)$

$Acceleration, \ a = \dfrac{dv}{dt}$

Therefore;

$\dfrac{dv}{dt} = b \cdot \left(\dfrac{g}{b} - v \right)$

The above differential equation gives;

$\dfrac{dv}{ \left(\dfrac{g}{b} - v \right)} = b \cdot dt$

Which gives;

$\displaystyle \int\limits {\dfrac{dv}{ \left(\dfrac{g}{b} - v \right)} } = \int\limits {b \cdot dt} = b \cdot t + C$

$\displaystyle \int\limits {\dfrac{dv}{ \left(\dfrac{g}{b} - v \right)} } = -\ln \left(\dfrac{g}{b} - v \right)$ and  $\displaystyle\int\limits{b \cdot dt} = b \cdot t + C$

Therefore

$\displaystyle -\ln \left(\dfrac{g}{b} - v \right) =b \cdot t + C$

At t = 0, v = 0, therefore;

$\displaystyle -\ln \left(\dfrac{g}{b} - 0 \right) =b \times 0 + C$

$C = \displaystyle -\ln \left(\dfrac{g}{b} \right)$

Which gives;

$\displaystyle -\ln \left(\dfrac{g}{b} - v \right) =b \cdot t \displaystyle -\ln \left(\dfrac{g}{b} \right)$

$\displaystyle \ln \left(\dfrac{g}{b} - v \right) =-b \cdot t \displaystyle +\ln \left(\dfrac{g}{b} \right)$

$\displaystyle \dfrac{g}{b} - v = e^{-b \cdot t \displaystyle +\ln \left(\dfrac{g}{b} \right)} = e^{-b \cdot t} \times e^\ln \left(\dfrac{g}{b} \right)} = e^{-b \cdot t} \times \dfrac{g}{b}$

$\displaystyle \dfrac{g}{b} - e^{-b \cdot t} \times \dfrac{g}{b} = v$

$\displaystyle \dfrac{g}{b} \cdot \left(1 - e^{-b \cdot t} \right) = v$

∴ v = g·(1 - $e^{-b \cdot t$)/b

The correct option is option (A)

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