A particle moving along a line has the position s(t) = t* – 162t² at time t seconds. What is the farthest distance to the left of the origin attained by the particle? (Give your answer as a whole or exact number.) S = m

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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### Problem Statement:
A particle moving along a line has the position \( s(t) = t^4 - 16t^2 \) at time \( t \) seconds. What is the farthest distance to the left of the origin attained by the particle?

(Give your answer as a whole or exact number.)

\[ s = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\ \text{m} \]

### Explanation:
To determine the farthest distance to the left of the origin, we need to find the minimum value of the position function \( s(t) \). Follow these steps:

1. **Find the First Derivative:**
   \[ s'(t) = \frac{d}{dt} (t^4 - 16t^2) = 4t^3 - 32t \]

2. **Set the First Derivative to Zero to Find Critical Points:**
   \[ 4t^3 - 32t = 0 \]
   \[ 4t(t^2 - 8) = 0 \]
   \[ 4t(t - 2\sqrt{2})(t + 2\sqrt{2}) = 0 \]
   \[ t = 0, \ \ t = 2\sqrt{2}, \ \ t = -2\sqrt{2} \]

3. **Find the Second Derivative to Determine Concavity:**
   \[ s''(t) = 12t^2 - 32 \]

   Evaluate the second derivative at each critical point:
   - For \( t = 0 \):
     \[ s''(0) = 12(0)^2 - 32 = -32 \ (\text{negative value indicates a local maximum}) \]
   - For \( t = 2\sqrt{2} \):
     \[ s''(2\sqrt{2}) = 12(8) - 32 = 96 - 32 = 64 \ (\text{positive value indicates a local minimum}) \]
   - For \( t = -2\sqrt{2} \):
     \[ s''(-2\sqrt{2}) = 12(8) - 32 = 96 - 32 = 64 \ (\text{positive value indicates a local minimum}) \]

4. **Evaluate
Transcribed Image Text:### Problem Statement: A particle moving along a line has the position \( s(t) = t^4 - 16t^2 \) at time \( t \) seconds. What is the farthest distance to the left of the origin attained by the particle? (Give your answer as a whole or exact number.) \[ s = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\ \text{m} \] ### Explanation: To determine the farthest distance to the left of the origin, we need to find the minimum value of the position function \( s(t) \). Follow these steps: 1. **Find the First Derivative:** \[ s'(t) = \frac{d}{dt} (t^4 - 16t^2) = 4t^3 - 32t \] 2. **Set the First Derivative to Zero to Find Critical Points:** \[ 4t^3 - 32t = 0 \] \[ 4t(t^2 - 8) = 0 \] \[ 4t(t - 2\sqrt{2})(t + 2\sqrt{2}) = 0 \] \[ t = 0, \ \ t = 2\sqrt{2}, \ \ t = -2\sqrt{2} \] 3. **Find the Second Derivative to Determine Concavity:** \[ s''(t) = 12t^2 - 32 \] Evaluate the second derivative at each critical point: - For \( t = 0 \): \[ s''(0) = 12(0)^2 - 32 = -32 \ (\text{negative value indicates a local maximum}) \] - For \( t = 2\sqrt{2} \): \[ s''(2\sqrt{2}) = 12(8) - 32 = 96 - 32 = 64 \ (\text{positive value indicates a local minimum}) \] - For \( t = -2\sqrt{2} \): \[ s''(-2\sqrt{2}) = 12(8) - 32 = 96 - 32 = 64 \ (\text{positive value indicates a local minimum}) \] 4. **Evaluate
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