Substance A decomposes at a rate proportional to the amount of A present. It is found that 18 lb of A will reduce to 9 lb in 4.3 hr. After how long will there be only 1 lb left? (...) There will be 1 lb left after hr. (Do not round until the final answer. Then round to the nearest whole number as needed.)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**Problem:**

Substance A decomposes at a rate proportional to the amount of A present. It is found that 18 lb of A will reduce to 9 lb in 4.3 hr. After how long will there be only 1 lb left?

---

**Answer Field:**

There will be 1 lb left after [   ] hr.

*(Do not round until the final answer. Then round to the nearest whole number as needed.)*

---

**Explanation:**

This problem involves exponential decay, where the rate of decomposition is proportional to the current amount. This implies the use of the exponential decay formula:

\[ A(t) = A_0 \cdot e^{-kt} \]

where:
- \( A(t) \) is the amount of substance at time \( t \),
- \( A_0 \) is the initial amount of the substance,
- \( k \) is the decay constant,
- \( t \) is the time.

Given:
- \( A_0 = 18 \) lb,
- \( A(t) = 9 \) lb when \( t = 4.3 \) hr.

Using this information, we first determine the decay constant \( k \) and then solve for the time when \( A(t) = 1 \) lb.
Transcribed Image Text:**Problem:** Substance A decomposes at a rate proportional to the amount of A present. It is found that 18 lb of A will reduce to 9 lb in 4.3 hr. After how long will there be only 1 lb left? --- **Answer Field:** There will be 1 lb left after [ ] hr. *(Do not round until the final answer. Then round to the nearest whole number as needed.)* --- **Explanation:** This problem involves exponential decay, where the rate of decomposition is proportional to the current amount. This implies the use of the exponential decay formula: \[ A(t) = A_0 \cdot e^{-kt} \] where: - \( A(t) \) is the amount of substance at time \( t \), - \( A_0 \) is the initial amount of the substance, - \( k \) is the decay constant, - \( t \) is the time. Given: - \( A_0 = 18 \) lb, - \( A(t) = 9 \) lb when \( t = 4.3 \) hr. Using this information, we first determine the decay constant \( k \) and then solve for the time when \( A(t) = 1 \) lb.
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