In a later chapter we will be able to show, under certain assumptions, that the velocity v(t) of a falling raindrop at time t is

Question

1 month ago

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v(t) = vT(1 − e−gt/vT) where g is the acceleration due to gravity and vT is the terminal velocity of the raindrop. (a) Find lim t→[infinity] v(t).

1 answer:

Maru [3.3K] · Answer 1 · 2026-02-20T17:50:25+03:00

Answer:

$\lim_{t \to \infty} v(t) =vT$

Explanation:

Applying the distributive property:

$v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}$

Thus:

$\lim_{t\to \infty} vT-e^{-gt}$

The sum of the limits of two functions equals the limit of their sum, therefore:

$\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}$

For a constant function, the limit remains the same constant, hence:

$\lim_{t\to \infty} vT=vT$

Next, we will evaluate the other limit:

$\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}$

The limit for a constant multiplied by a function equals the product of the constant and the limit of the function, thus:

$\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)$

$-g(\infty)=-\infty$

Consequently:

$e^{(-\infty)} =0$

In conclusion:

$\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT$