Answer :
To determine when the particle reaches its minimum velocity, we can follow these steps:
1. Define the position function:
The position of the particle is given by:
[tex]\[ x(t) = 2t^3 - 8t^2 + 12 \][/tex]
2. Find the velocity function:
The velocity is the first derivative of the position function with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 8t^2 + 12) \][/tex]
Calculating this derivative, we get:
[tex]\[ v(t) = 6t^2 - 16t \][/tex]
3. Find the acceleration function:
The acceleration is the first derivative of the velocity function with respect to time [tex]\( t \)[/tex]:
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 16t) \][/tex]
Calculating this derivative, we get:
[tex]\[ a(t) = 12t - 16 \][/tex]
4. Determine when the acceleration is zero:
To find the time when the velocity is at a minimum, we set the acceleration function to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 12t - 16 = 0 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ t = \frac{16}{12} = \frac{4}{3} \][/tex]
5. Verify that the velocity at this time is a minimum:
We use the second derivative test to confirm that we have a minimum. The second derivative of acceleration (or the third derivative of the position function) is:
[tex]\[ a''(t) = \frac{d^2v}{dt^2} = \frac{d}{dt}(12t - 16) = 12 \][/tex]
Since [tex]\( a''(t) = 12 \)[/tex] is a positive constant, [tex]\( t = \frac{4}{3} \)[/tex] is indeed a minimum.
Thus, the particle reaches its minimum velocity at [tex]\( t = \frac{4}{3} \)[/tex] seconds.
1. Define the position function:
The position of the particle is given by:
[tex]\[ x(t) = 2t^3 - 8t^2 + 12 \][/tex]
2. Find the velocity function:
The velocity is the first derivative of the position function with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 8t^2 + 12) \][/tex]
Calculating this derivative, we get:
[tex]\[ v(t) = 6t^2 - 16t \][/tex]
3. Find the acceleration function:
The acceleration is the first derivative of the velocity function with respect to time [tex]\( t \)[/tex]:
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 16t) \][/tex]
Calculating this derivative, we get:
[tex]\[ a(t) = 12t - 16 \][/tex]
4. Determine when the acceleration is zero:
To find the time when the velocity is at a minimum, we set the acceleration function to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 12t - 16 = 0 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ t = \frac{16}{12} = \frac{4}{3} \][/tex]
5. Verify that the velocity at this time is a minimum:
We use the second derivative test to confirm that we have a minimum. The second derivative of acceleration (or the third derivative of the position function) is:
[tex]\[ a''(t) = \frac{d^2v}{dt^2} = \frac{d}{dt}(12t - 16) = 12 \][/tex]
Since [tex]\( a''(t) = 12 \)[/tex] is a positive constant, [tex]\( t = \frac{4}{3} \)[/tex] is indeed a minimum.
Thus, the particle reaches its minimum velocity at [tex]\( t = \frac{4}{3} \)[/tex] seconds.
