Answer :
Sure, let's walk through each part of the question step-by-step.
### Part (a): Deriving Velocity and Acceleration
The position of the object is given by the function [tex]\( x(t) = 3 - 2t^2 + 3t^3 \)[/tex].
1. Velocity is the first derivative of the position function with respect to time.
[tex]\[ v(t) = \frac{dx}{dt} = \frac{d}{dt} (3 - 2t^2 + 3t^3) = -4t + 9t^2 \][/tex]
2. Acceleration is the derivative of the velocity function with respect to time.
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt} (-4t + 9t^2) = -4 + 18t \][/tex]
### Part (b): Position of the Object at Specific Times
1. At [tex]\( t = 0 \)[/tex]:
[tex]\[ x(0) = 3 - 2(0)^2 + 3(0)^3 = 3 \][/tex]
2. At [tex]\( t = 2 \)[/tex]:
[tex]\[ x(2) = 3 - 2(2)^2 + 3(2)^3 = 3 - 8 + 24 = 19 \][/tex]
3. At [tex]\( t = 4 \)[/tex]:
[tex]\[ x(4) = 3 - 2(4)^2 + 3(4)^3 = 3 - 32 + 192 = 163 \][/tex]
### Part (c): Displacement Between Time Intervals
1. Between [tex]\( t = 2 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Displacement} = x(4) - x(2) = 163 - 19 = 144 \][/tex]
2. Between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Displacement} = x(4) - x(0) = 163 - 3 = 160 \][/tex]
### Part (d): Average Velocity
1. Between [tex]\( t = 2 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{x(4) - x(2)}{4 - 2} = \frac{163 - 19}{2} = \frac{144}{2} = 72.0 \, \text{m/s} \][/tex]
2. Between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{x(4) - x(0)}{4 - 0} = \frac{163 - 3}{4} = \frac{160}{4} = 40.0 \, \text{m/s} \][/tex]
3. Between [tex]\( t = 1 \)[/tex] and [tex]\( t = 3 \)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{x(3) - x(1)}{3 - 1} = \frac{77 - 15}{2} = \frac{62}{2} = 31.0 \, \text{m/s} \][/tex]
### Part (e): Instantaneous Velocity
1. At [tex]\( t = 2 \)[/tex]:
[tex]\[ v(2) = -4(2) + 9(2)^2 = -8 + 36 = 28 \, \text{m/s} \][/tex]
2. At [tex]\( t = 5 \)[/tex]:
[tex]\[ v(5) = -4(5) + 9(5)^2 = -20 + 225 = 205 \, \text{m/s} \][/tex]
### Part (f): Instantaneous Velocity Zero Times
To find when the instantaneous velocity is zero, we solve the equation [tex]\( v(t) = 0 \)[/tex]:
[tex]\[ -4t + 9t^2 = 0 \][/tex]
[tex]\[ t(-4 + 9t) = 0 \][/tex]
This gives us:
[tex]\[ t = 0 \quad \text{or} \quad 9t = 4 \quad \Rightarrow \quad t = \frac{4}{9} \quad \approx \quad 0.4444 \][/tex]
Thus, the instantaneous velocity is zero at [tex]\( t = 0 \)[/tex] and [tex]\( t \approx 0.444 \)[/tex].
These results encapsulate the detailed solutions for each part of the question.
### Part (a): Deriving Velocity and Acceleration
The position of the object is given by the function [tex]\( x(t) = 3 - 2t^2 + 3t^3 \)[/tex].
1. Velocity is the first derivative of the position function with respect to time.
[tex]\[ v(t) = \frac{dx}{dt} = \frac{d}{dt} (3 - 2t^2 + 3t^3) = -4t + 9t^2 \][/tex]
2. Acceleration is the derivative of the velocity function with respect to time.
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt} (-4t + 9t^2) = -4 + 18t \][/tex]
### Part (b): Position of the Object at Specific Times
1. At [tex]\( t = 0 \)[/tex]:
[tex]\[ x(0) = 3 - 2(0)^2 + 3(0)^3 = 3 \][/tex]
2. At [tex]\( t = 2 \)[/tex]:
[tex]\[ x(2) = 3 - 2(2)^2 + 3(2)^3 = 3 - 8 + 24 = 19 \][/tex]
3. At [tex]\( t = 4 \)[/tex]:
[tex]\[ x(4) = 3 - 2(4)^2 + 3(4)^3 = 3 - 32 + 192 = 163 \][/tex]
### Part (c): Displacement Between Time Intervals
1. Between [tex]\( t = 2 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Displacement} = x(4) - x(2) = 163 - 19 = 144 \][/tex]
2. Between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Displacement} = x(4) - x(0) = 163 - 3 = 160 \][/tex]
### Part (d): Average Velocity
1. Between [tex]\( t = 2 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{x(4) - x(2)}{4 - 2} = \frac{163 - 19}{2} = \frac{144}{2} = 72.0 \, \text{m/s} \][/tex]
2. Between [tex]\( t = 0 \)[/tex] and [tex]\( t = 4 \)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{x(4) - x(0)}{4 - 0} = \frac{163 - 3}{4} = \frac{160}{4} = 40.0 \, \text{m/s} \][/tex]
3. Between [tex]\( t = 1 \)[/tex] and [tex]\( t = 3 \)[/tex]:
[tex]\[ \text{Average Velocity} = \frac{x(3) - x(1)}{3 - 1} = \frac{77 - 15}{2} = \frac{62}{2} = 31.0 \, \text{m/s} \][/tex]
### Part (e): Instantaneous Velocity
1. At [tex]\( t = 2 \)[/tex]:
[tex]\[ v(2) = -4(2) + 9(2)^2 = -8 + 36 = 28 \, \text{m/s} \][/tex]
2. At [tex]\( t = 5 \)[/tex]:
[tex]\[ v(5) = -4(5) + 9(5)^2 = -20 + 225 = 205 \, \text{m/s} \][/tex]
### Part (f): Instantaneous Velocity Zero Times
To find when the instantaneous velocity is zero, we solve the equation [tex]\( v(t) = 0 \)[/tex]:
[tex]\[ -4t + 9t^2 = 0 \][/tex]
[tex]\[ t(-4 + 9t) = 0 \][/tex]
This gives us:
[tex]\[ t = 0 \quad \text{or} \quad 9t = 4 \quad \Rightarrow \quad t = \frac{4}{9} \quad \approx \quad 0.4444 \][/tex]
Thus, the instantaneous velocity is zero at [tex]\( t = 0 \)[/tex] and [tex]\( t \approx 0.444 \)[/tex].
These results encapsulate the detailed solutions for each part of the question.
