Answer :
Let's break down the problem step by step.
1. Finding the velocity function
The position of the particle is given by:
[tex]\[ s(t) = 2t^3 - 21t^2 + 36t \][/tex]
The velocity of the particle is the derivative of the position function with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}[2t^3 - 21t^2 + 36t] \][/tex]
Differentiating term by term:
[tex]\[ v(t) = 6t^2 - 42t + 36 \][/tex]
2. Finding the velocity at time [tex]\( t = 0 \)[/tex]
Substitute [tex]\( t = 0 \)[/tex] into the velocity function:
[tex]\[ v(0) = 6(0)^2 - 42(0) + 36 = 36 \text{ ft/sec} \][/tex]
So, the velocity of the particle at time [tex]\( t = 0 \)[/tex] is [tex]\( \boxed{36} \)[/tex] ft/sec.
3. Finding the times when the particle is at rest
The particle is at rest when the velocity is zero:
[tex]\[ v(t) = 6t^2 - 42t + 36 = 0 \][/tex]
Solving the quadratic equation:
[tex]\[ 6t^2 - 42t + 36 = 0 \][/tex]
First, simplify by dividing through by 6:
[tex]\[ t^2 - 7t + 6 = 0 \][/tex]
Factoring the quadratic equation:
[tex]\[ (t - 1)(t - 6) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ t = 1 \quad \text{and} \quad t = 6 \][/tex]
So, the particle stops moving at [tex]\( t = \boxed{1} \)[/tex] and [tex]\( t = \boxed{6} \)[/tex].
4. Finding the position of the particle at time [tex]\( t = 14 \)[/tex]
Substitute [tex]\( t = 14 \)[/tex] into the position function:
[tex]\[ s(14) = 2(14)^3 - 21(14)^2 + 36(14) \][/tex]
Calculating step by step:
[tex]\[ 14^3 = 2744, \quad 14^2 = 196 \][/tex]
[tex]\[ s(14) = 2 \times 2744 - 21 \times 196 + 36 \times 14 \][/tex]
[tex]\[ = 5488 - 4116 + 504 \][/tex]
[tex]\[ = 1876 \text{ feet} \][/tex]
So, the position of the particle at time [tex]\( 14 \)[/tex] is [tex]\( \boxed{1876} \)[/tex] feet.
5. Finding the total distance traveled by the particle between time [tex]\( 0 \)[/tex] and [tex]\( 14 \)[/tex]
To find the total distance traveled, we need to integrate the absolute value of the velocity function from [tex]\( t = 0 \)[/tex] to [tex]\( t = 14 \)[/tex]:
The total distance is given by:
[tex]\[ \int_{0}^{14} |v(t)| \, dt \][/tex]
Using the results, the total distance traveled by the particle is:
[tex]\[ \boxed{2126} \text{ feet} \][/tex]
1. Finding the velocity function
The position of the particle is given by:
[tex]\[ s(t) = 2t^3 - 21t^2 + 36t \][/tex]
The velocity of the particle is the derivative of the position function with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}[2t^3 - 21t^2 + 36t] \][/tex]
Differentiating term by term:
[tex]\[ v(t) = 6t^2 - 42t + 36 \][/tex]
2. Finding the velocity at time [tex]\( t = 0 \)[/tex]
Substitute [tex]\( t = 0 \)[/tex] into the velocity function:
[tex]\[ v(0) = 6(0)^2 - 42(0) + 36 = 36 \text{ ft/sec} \][/tex]
So, the velocity of the particle at time [tex]\( t = 0 \)[/tex] is [tex]\( \boxed{36} \)[/tex] ft/sec.
3. Finding the times when the particle is at rest
The particle is at rest when the velocity is zero:
[tex]\[ v(t) = 6t^2 - 42t + 36 = 0 \][/tex]
Solving the quadratic equation:
[tex]\[ 6t^2 - 42t + 36 = 0 \][/tex]
First, simplify by dividing through by 6:
[tex]\[ t^2 - 7t + 6 = 0 \][/tex]
Factoring the quadratic equation:
[tex]\[ (t - 1)(t - 6) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ t = 1 \quad \text{and} \quad t = 6 \][/tex]
So, the particle stops moving at [tex]\( t = \boxed{1} \)[/tex] and [tex]\( t = \boxed{6} \)[/tex].
4. Finding the position of the particle at time [tex]\( t = 14 \)[/tex]
Substitute [tex]\( t = 14 \)[/tex] into the position function:
[tex]\[ s(14) = 2(14)^3 - 21(14)^2 + 36(14) \][/tex]
Calculating step by step:
[tex]\[ 14^3 = 2744, \quad 14^2 = 196 \][/tex]
[tex]\[ s(14) = 2 \times 2744 - 21 \times 196 + 36 \times 14 \][/tex]
[tex]\[ = 5488 - 4116 + 504 \][/tex]
[tex]\[ = 1876 \text{ feet} \][/tex]
So, the position of the particle at time [tex]\( 14 \)[/tex] is [tex]\( \boxed{1876} \)[/tex] feet.
5. Finding the total distance traveled by the particle between time [tex]\( 0 \)[/tex] and [tex]\( 14 \)[/tex]
To find the total distance traveled, we need to integrate the absolute value of the velocity function from [tex]\( t = 0 \)[/tex] to [tex]\( t = 14 \)[/tex]:
The total distance is given by:
[tex]\[ \int_{0}^{14} |v(t)| \, dt \][/tex]
Using the results, the total distance traveled by the particle is:
[tex]\[ \boxed{2126} \text{ feet} \][/tex]
