Answer :
To solve the given problem, we need to perform several steps involving calculus to find the velocity and acceleration of the particle at specific times. Let's break this down step by step.
### Step 1: Determine the Velocity Function
The velocity of the particle is given by the first derivative of the displacement function [tex]\( S \)[/tex] with respect to time [tex]\( t \)[/tex].
The displacement function is:
[tex]\[ S(t) = 4t - 2t^2 - 5t^3 \][/tex]
To find the velocity, we differentiate [tex]\( S \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{dS}{dt} = \frac{d}{dt} (4t - 2t^2 - 5t^3) = 4 - 4t - 15t^2 \][/tex]
### Step 2: Find the Velocity at [tex]\( t = 38 \)[/tex]
To determine the velocity at [tex]\( t = 38 \)[/tex], we substitute [tex]\( t = 38 \)[/tex] into the velocity function [tex]\( v(t) \)[/tex]:
[tex]\[ v(38) = 4 - 4(38) - 15(38^2) \][/tex]
Performing the arithmetic:
[tex]\[ v(38) = 4 - 152 - 15(1444) \][/tex]
[tex]\[ v(38) = 4 - 152 - 21660 \][/tex]
[tex]\[ v(38) = 4 - 21812 \][/tex]
[tex]\[ v(38) = -21808 \][/tex]
So, the velocity at [tex]\( t = 38 \)[/tex] is [tex]\( -21808 \)[/tex].
### Step 3: Find the Time When Velocity is Zero
To find the time [tex]\( t \)[/tex] when the velocity is zero, we set the velocity function equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 4 - 4t - 15t^2 = 0 \][/tex]
This is a quadratic equation in the form:
[tex]\[ -15t^2 - 4t + 4 = 0 \][/tex]
Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = -15 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 4 \)[/tex].
The solutions to the equation are:
[tex]\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-15)(4)}}{2(-15)} \][/tex]
[tex]\[ t = \frac{4 \pm \sqrt{16 + 240}}{-30} \][/tex]
[tex]\[ t = \frac{4 \pm \sqrt{256}}{-30} \][/tex]
[tex]\[ t = \frac{4 \pm 16}{-30} \][/tex]
Thus, we have two solutions:
[tex]\[ t = \frac{4 + 16}{-30} = \frac{20}{-30} = -\frac{2}{3} \][/tex]
[tex]\[ t = \frac{4 - 16}{-30} = \frac{-12}{-30} = \frac{2}{5} \][/tex]
So, the times when the velocity is zero are [tex]\( t = -\frac{2}{3} \)[/tex] and [tex]\( t = \frac{2}{5} \)[/tex].
### Step 4: Determine the Acceleration Function
The acceleration of the particle is given by the second derivative of the displacement function [tex]\( S \)[/tex] with respect to time [tex]\( t \)[/tex].
We already have the first derivative (velocity function):
[tex]\[ v(t) = 4 - 4t - 15t^2 \][/tex]
To find the acceleration, we differentiate the velocity function [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt} (4 - 4t - 15t^2) = -4 - 30t \][/tex]
### Step 5: Find the Acceleration at [tex]\( t = 2 \)[/tex]
To determine the acceleration at [tex]\( t = 2 \)[/tex], we substitute [tex]\( t = 2 \)[/tex] into the acceleration function [tex]\( a(t) \)[/tex]:
[tex]\[ a(2) = -4 - 30(2) \][/tex]
[tex]\[ a(2) = -4 - 60 \][/tex]
[tex]\[ a(2) = -64 \][/tex]
So, the acceleration at [tex]\( t = 2 \)[/tex] is [tex]\( -64 \)[/tex].
### Summary
1. The velocity at [tex]\( t = 38 \)[/tex] is [tex]\( -21808 \)[/tex].
2. The times when the velocity is zero are [tex]\( t = -\frac{2}{3} \)[/tex] and [tex]\( t = \frac{2}{5} \)[/tex].
3. The acceleration at [tex]\( t = 2 \)[/tex] is [tex]\( -64 \)[/tex].
### Step 1: Determine the Velocity Function
The velocity of the particle is given by the first derivative of the displacement function [tex]\( S \)[/tex] with respect to time [tex]\( t \)[/tex].
The displacement function is:
[tex]\[ S(t) = 4t - 2t^2 - 5t^3 \][/tex]
To find the velocity, we differentiate [tex]\( S \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{dS}{dt} = \frac{d}{dt} (4t - 2t^2 - 5t^3) = 4 - 4t - 15t^2 \][/tex]
### Step 2: Find the Velocity at [tex]\( t = 38 \)[/tex]
To determine the velocity at [tex]\( t = 38 \)[/tex], we substitute [tex]\( t = 38 \)[/tex] into the velocity function [tex]\( v(t) \)[/tex]:
[tex]\[ v(38) = 4 - 4(38) - 15(38^2) \][/tex]
Performing the arithmetic:
[tex]\[ v(38) = 4 - 152 - 15(1444) \][/tex]
[tex]\[ v(38) = 4 - 152 - 21660 \][/tex]
[tex]\[ v(38) = 4 - 21812 \][/tex]
[tex]\[ v(38) = -21808 \][/tex]
So, the velocity at [tex]\( t = 38 \)[/tex] is [tex]\( -21808 \)[/tex].
### Step 3: Find the Time When Velocity is Zero
To find the time [tex]\( t \)[/tex] when the velocity is zero, we set the velocity function equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ 4 - 4t - 15t^2 = 0 \][/tex]
This is a quadratic equation in the form:
[tex]\[ -15t^2 - 4t + 4 = 0 \][/tex]
Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = -15 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 4 \)[/tex].
The solutions to the equation are:
[tex]\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-15)(4)}}{2(-15)} \][/tex]
[tex]\[ t = \frac{4 \pm \sqrt{16 + 240}}{-30} \][/tex]
[tex]\[ t = \frac{4 \pm \sqrt{256}}{-30} \][/tex]
[tex]\[ t = \frac{4 \pm 16}{-30} \][/tex]
Thus, we have two solutions:
[tex]\[ t = \frac{4 + 16}{-30} = \frac{20}{-30} = -\frac{2}{3} \][/tex]
[tex]\[ t = \frac{4 - 16}{-30} = \frac{-12}{-30} = \frac{2}{5} \][/tex]
So, the times when the velocity is zero are [tex]\( t = -\frac{2}{3} \)[/tex] and [tex]\( t = \frac{2}{5} \)[/tex].
### Step 4: Determine the Acceleration Function
The acceleration of the particle is given by the second derivative of the displacement function [tex]\( S \)[/tex] with respect to time [tex]\( t \)[/tex].
We already have the first derivative (velocity function):
[tex]\[ v(t) = 4 - 4t - 15t^2 \][/tex]
To find the acceleration, we differentiate the velocity function [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt} (4 - 4t - 15t^2) = -4 - 30t \][/tex]
### Step 5: Find the Acceleration at [tex]\( t = 2 \)[/tex]
To determine the acceleration at [tex]\( t = 2 \)[/tex], we substitute [tex]\( t = 2 \)[/tex] into the acceleration function [tex]\( a(t) \)[/tex]:
[tex]\[ a(2) = -4 - 30(2) \][/tex]
[tex]\[ a(2) = -4 - 60 \][/tex]
[tex]\[ a(2) = -64 \][/tex]
So, the acceleration at [tex]\( t = 2 \)[/tex] is [tex]\( -64 \)[/tex].
### Summary
1. The velocity at [tex]\( t = 38 \)[/tex] is [tex]\( -21808 \)[/tex].
2. The times when the velocity is zero are [tex]\( t = -\frac{2}{3} \)[/tex] and [tex]\( t = \frac{2}{5} \)[/tex].
3. The acceleration at [tex]\( t = 2 \)[/tex] is [tex]\( -64 \)[/tex].
