Answer :
To determine the acceleration at [tex]\( t = 2 \)[/tex] for the given position function [tex]\( s(t) = 2t^3 + 2t^2 + 5t - 20 \, \text{m} \)[/tex], we need to follow these steps:
1. Find the Velocity Function:
The velocity function is the first derivative of the position function with respect to time [tex]\( t \)[/tex].
[tex]\[ v(t) = \frac{d}{dt}[2t^3 + 2t^2 + 5t - 20] \][/tex]
Taking the derivative:
[tex]\[ v(t) = 6t^2 + 4t + 5 \, \text{m/s} \][/tex]
2. Find the Acceleration Function:
The acceleration function is the first derivative of the velocity function, which means it is the second derivative of the position function.
[tex]\[ a(t) = \frac{d}{dt}[6t^2 + 4t + 5] \][/tex]
Taking the derivative:
[tex]\[ a(t) = 12t + 4 \, \text{m/s}^2 \][/tex]
3. Evaluate the Acceleration at [tex]\( t = 2 \)[/tex]:
Substitute [tex]\( t = 2 \)[/tex] into the acceleration function:
[tex]\[ a(2) = 12(2) + 4 \][/tex]
Simplifying this:
[tex]\[ a(2) = 24 + 4 = 28 \, \text{m/s}^2 \][/tex]
Thus, the acceleration at [tex]\( t = 2 \)[/tex] is [tex]\( 28 \, \text{m/s}^2 \)[/tex].
Based on the provided options, the correct answer is:
(C) [tex]\( 28 \, \text{m/s}^2 \)[/tex]
1. Find the Velocity Function:
The velocity function is the first derivative of the position function with respect to time [tex]\( t \)[/tex].
[tex]\[ v(t) = \frac{d}{dt}[2t^3 + 2t^2 + 5t - 20] \][/tex]
Taking the derivative:
[tex]\[ v(t) = 6t^2 + 4t + 5 \, \text{m/s} \][/tex]
2. Find the Acceleration Function:
The acceleration function is the first derivative of the velocity function, which means it is the second derivative of the position function.
[tex]\[ a(t) = \frac{d}{dt}[6t^2 + 4t + 5] \][/tex]
Taking the derivative:
[tex]\[ a(t) = 12t + 4 \, \text{m/s}^2 \][/tex]
3. Evaluate the Acceleration at [tex]\( t = 2 \)[/tex]:
Substitute [tex]\( t = 2 \)[/tex] into the acceleration function:
[tex]\[ a(2) = 12(2) + 4 \][/tex]
Simplifying this:
[tex]\[ a(2) = 24 + 4 = 28 \, \text{m/s}^2 \][/tex]
Thus, the acceleration at [tex]\( t = 2 \)[/tex] is [tex]\( 28 \, \text{m/s}^2 \)[/tex].
Based on the provided options, the correct answer is:
(C) [tex]\( 28 \, \text{m/s}^2 \)[/tex]