Mathematics, 20.06.2020 03:57 ajayrose

Substance A decomposes at a rate proportional to the amount of A present. a) Write an equation relating A to the amount left of an initial amount Upper A 0 after time t. b) It is found that 18 lb of A will reduce to 9 lb in 4.2 hr. After how long will there be only 1 lb left?

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Answer from: lilquongohard

The equation relating A to the amount left of an initial amount $\rm A_0$ after time t is $\rm A(t) = A_0.e^{kt}$ .

After 17.5 hours there be only 1 lb left.

Given that,

Substance A decomposes at a rate proportional to the amount of A present.

We have to find,

Write an equation relating A to the amount left of an initial amount Upper A 0 after time t.

It is found that 18 lb of A will reduce to 9 lb in 4.2 hr. After how long will there be only 1 lb left?

According to the question,

1. Substance A decomposes at a rate proportional to the amount of A present.

$\rm A(t) = A_0.e^{kt}$

Here, $\rm A_0$ is the amount of A present before decomposition.

A is the amount left after decomposition after time t.

The equation relating A to the amount left of an initial amount $\rm A_0$ after time t is $\rm A(t) = A_0.e^{kt}$ .

2. The amount of substance 18 lb of A will reduce to 9 lb in 4.2 hours.

Here, the amount of A present before decomposition $\rm A_0$ is 18 lb.

And A is the amount left is 9 lb after decomposition after time t is 4.2 hours.

Therefore,

$\rm A(t) = A_0.e^{kt}\\\\\rm 9=18\times e^{(4.2k)} \\\\\dfrac{9}{18} = e^{(4.2k)}\\\\ \dfrac{1}{2} = e^{(4.2k)}\\\\ \ 0.5= e^{(4.2k)}\\\\ Taking \ log \ on \ both \ sides\\\\log(0.5 )= 4.2k \\\\-0.693 = 4.2k\\\\k = \dfrac{-0.693}{4}\\\\k = 0.165$

The time when only 1 lb left is,

$\rm A = A_0 e ^{(.kt)}\\\\1 = 18 \times e^{-0.165t}\\\\\dfrac{1}{18}= e^{-0.165t}\\\\0.056 = e^{-0.165t}\\\\ Taking \ log \ on \ both \ sides \\\\log(0.056 )= {-0.165t}\\\\-2.88 = {-0.165t}\\\\t = \dfrac{-2.88}{-0.165}\\\\t = 17.5 \ hours$

Hence, After 17.5 hours there be only 1 lb left.

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Answer from: sophiaa23

Step-by-step explanation:

a) The equation relating A to the amount left of an initial amount Upper A 0 after time t is

A(t) = A0e^(kt)

b) It is found that 18 lb of A will reduce to 9 lb in 4.2 hr. It means that

A0 = 18

A = 9

t = 4.2

Therefore, substituting into the formula, it becomes

9 = 18e^(4.2k)

9/18 = e^(4.2k)

0.5 = e^(4.2k)

Taking ln of both sides of the equation, it becomes

Ln0.5 = ln e^(4.2k)

- 0.693 = 4.2k

k = - 0.693/4.2

k = - 0.165

The equation becomes

A(t) = 18e^(- 0.165t)

For A = 1(1 lb left), then

1 = 18e^(- 0.165t)

1/18 = e^(- 0.165t)

0.056 = e^(- 0.165t)

Taking ln of both sides, it becomes

Ln0.056 = ln e^(- 0.165t)

- 2.88 = - 0.165t

t = - 2.88/-0.165

t = 17.5 hours

After 17.5 hours, 1 lb would be left

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