Answer :
Sure! Let's solve the given problem step-by-step, analyzing the motion of the particle described by the displacement function [tex]\( s(t) = -31t^2 + 21t + 5 \)[/tex], where [tex]\( s \)[/tex] is the displacement in meters and [tex]\( t \)[/tex] is the time in seconds.
### Part 1: Velocity and Acceleration after 4 seconds
1. Displacement Function:
[tex]\[ s(t) = -31t^2 + 21t + 5 \][/tex]
2. Velocity is the first derivative of displacement with respect to time:
[tex]\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(-31t^2 + 21t + 5) \][/tex]
Applying differentiation, we get:
[tex]\[ v(t) = -62t + 21 \][/tex]
3. Acceleration is the first derivative of velocity with respect to time:
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-62t + 21) \][/tex]
Applying differentiation, we get:
[tex]\[ a(t) = -62 \][/tex]
4. Velocity after 4 seconds [tex]\((t = 4)\)[/tex]:
[tex]\[ v(4) = -62 \cdot 4 + 21 = -248 + 21 = -227 \, \text{m/s} \][/tex]
5. Acceleration after 4 seconds [tex]\((t = 4)\)[/tex]:
Since acceleration [tex]\(a(t)\)[/tex] is a constant, it remains:
[tex]\[ a(4) = -62 \, \text{m/s}^2 \][/tex]
Thus, the velocity after 4 seconds is [tex]\(-227 \, \text{m/s}\)[/tex] and the acceleration after 4 seconds is [tex]\(-62 \, \text{m/s}^2\)[/tex].
### Part 2: Maximum or Minimum Velocity and Corresponding Displacement
1. To find the maximum or minimum velocity, determine critical points where the derivative of velocity (the acceleration) is zero:
Since the acceleration is constant and nonzero [tex]\(-62 \, \text{m/s}^2\)[/tex], it does not have critical points, meaning it doesn’t change sign, and there are no local maxima or minima.
2. However, set the velocity equation to zero to find points where velocity is zero:
[tex]\[ -62t + 21 = 0 \implies t = \frac{21}{62} \][/tex]
At [tex]\( t = \frac{21}{62} \)[/tex] seconds, velocity is zero.
3. Displacement at this time:
[tex]\[ s\left( \frac{21}{62} \right) = -31 \left( \frac{21}{62} \right)^2 + 21 \left( \frac{21}{62} \right) + 5 \][/tex]
Simplifying, this results in:
[tex]\[ s\left( \frac{21}{62} \right) = \frac{1061}{124} \, \text{meters} \][/tex]
Thus, the velocity reaches zero at [tex]\( t = \frac{21}{62} \)[/tex] seconds, and the corresponding displacement is [tex]\(\frac{1061}{124}\)[/tex] meters.
### Part 3: Time at Which Velocity is Zero
1. From the previous part, we established that:
[tex]\[ v(t) = -62t + 21 \][/tex]
Setting [tex]\( v(t) \)[/tex] to zero for finding when velocity is zero:
[tex]\[ -62t + 21 = 0 \implies t = \frac{21}{62} \][/tex]
So, the time at which the velocity is zero is [tex]\( t = \frac{21}{62} \)[/tex] seconds.
### Summary:
- Velocity after 4 seconds: [tex]\(-227 \, \text{m/s}\)[/tex]
- Acceleration after 4 seconds: [tex]\(-62 \, \text{m/s}^2\)[/tex]
- Velocity at critical points: [tex]\(0 \, \text{m/s}\)[/tex] at [tex]\( t = \frac{21}{62} \, \text{seconds}\)[/tex]
- Displacement at critical points: [tex]\( \frac{1061}{124} \, \text{meters}\)[/tex]
- Time at which velocity is zero: [tex]\( t = \frac{21}{62} \, \text{seconds} \)[/tex]
These are the results for the given problem.
### Part 1: Velocity and Acceleration after 4 seconds
1. Displacement Function:
[tex]\[ s(t) = -31t^2 + 21t + 5 \][/tex]
2. Velocity is the first derivative of displacement with respect to time:
[tex]\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(-31t^2 + 21t + 5) \][/tex]
Applying differentiation, we get:
[tex]\[ v(t) = -62t + 21 \][/tex]
3. Acceleration is the first derivative of velocity with respect to time:
[tex]\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-62t + 21) \][/tex]
Applying differentiation, we get:
[tex]\[ a(t) = -62 \][/tex]
4. Velocity after 4 seconds [tex]\((t = 4)\)[/tex]:
[tex]\[ v(4) = -62 \cdot 4 + 21 = -248 + 21 = -227 \, \text{m/s} \][/tex]
5. Acceleration after 4 seconds [tex]\((t = 4)\)[/tex]:
Since acceleration [tex]\(a(t)\)[/tex] is a constant, it remains:
[tex]\[ a(4) = -62 \, \text{m/s}^2 \][/tex]
Thus, the velocity after 4 seconds is [tex]\(-227 \, \text{m/s}\)[/tex] and the acceleration after 4 seconds is [tex]\(-62 \, \text{m/s}^2\)[/tex].
### Part 2: Maximum or Minimum Velocity and Corresponding Displacement
1. To find the maximum or minimum velocity, determine critical points where the derivative of velocity (the acceleration) is zero:
Since the acceleration is constant and nonzero [tex]\(-62 \, \text{m/s}^2\)[/tex], it does not have critical points, meaning it doesn’t change sign, and there are no local maxima or minima.
2. However, set the velocity equation to zero to find points where velocity is zero:
[tex]\[ -62t + 21 = 0 \implies t = \frac{21}{62} \][/tex]
At [tex]\( t = \frac{21}{62} \)[/tex] seconds, velocity is zero.
3. Displacement at this time:
[tex]\[ s\left( \frac{21}{62} \right) = -31 \left( \frac{21}{62} \right)^2 + 21 \left( \frac{21}{62} \right) + 5 \][/tex]
Simplifying, this results in:
[tex]\[ s\left( \frac{21}{62} \right) = \frac{1061}{124} \, \text{meters} \][/tex]
Thus, the velocity reaches zero at [tex]\( t = \frac{21}{62} \)[/tex] seconds, and the corresponding displacement is [tex]\(\frac{1061}{124}\)[/tex] meters.
### Part 3: Time at Which Velocity is Zero
1. From the previous part, we established that:
[tex]\[ v(t) = -62t + 21 \][/tex]
Setting [tex]\( v(t) \)[/tex] to zero for finding when velocity is zero:
[tex]\[ -62t + 21 = 0 \implies t = \frac{21}{62} \][/tex]
So, the time at which the velocity is zero is [tex]\( t = \frac{21}{62} \)[/tex] seconds.
### Summary:
- Velocity after 4 seconds: [tex]\(-227 \, \text{m/s}\)[/tex]
- Acceleration after 4 seconds: [tex]\(-62 \, \text{m/s}^2\)[/tex]
- Velocity at critical points: [tex]\(0 \, \text{m/s}\)[/tex] at [tex]\( t = \frac{21}{62} \, \text{seconds}\)[/tex]
- Displacement at critical points: [tex]\( \frac{1061}{124} \, \text{meters}\)[/tex]
- Time at which velocity is zero: [tex]\( t = \frac{21}{62} \, \text{seconds} \)[/tex]
These are the results for the given problem.
