Answer :
Let's go through each part of the problem step-by-step.
### (A) Find the position of the body at [tex]\( t = 1 \)[/tex] s.
The position function is given by:
[tex]\[ x(t) = 16t - 6t^2 \][/tex]
Substitute [tex]\( t = 1 \)[/tex]:
[tex]\[ x(1) = 16(1) - 6(1)^2 \][/tex]
[tex]\[ x(1) = 16 - 6 \][/tex]
[tex]\[ x(1) = 10 \][/tex]
So, the position of the body at [tex]\( t = 1 \)[/tex] s is [tex]\( 10 \)[/tex] meters.
### (B) At what time does the body pass the origin?
To find when the body passes the origin, set [tex]\( x(t) = 0 \)[/tex]:
[tex]\[ 16t - 6t^2 = 0 \][/tex]
[tex]\[ 2t(8 - 3t) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ t = \frac{8}{3} = 2.67 \][/tex]
So, the body passes the origin at [tex]\( t = 0 \)[/tex] seconds and [tex]\( t = 2.67 \)[/tex] seconds.
### (C) Calculate the average velocity of the body between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex] seconds.
The average velocity is given by the change in position divided by the change in time:
[tex]\[ \text{Average Velocity} = \frac{x(2) - x(0)}{2 - 0} \][/tex]
First, find [tex]\( x(2) \)[/tex] and [tex]\( x(0) \)[/tex]:
[tex]\[ x(0) = 16(0) - 6(0)^2 = 0 \][/tex]
[tex]\[ x(2) = 16(2) - 6(2)^2 = 32 - 24 = 8 \][/tex]
So,
[tex]\[ \text{Average Velocity} = \frac{8 - 0}{2 - 0} = \frac{8}{2} = 4 \][/tex]
The average velocity between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex] seconds is [tex]\( 4 \)[/tex] meters per second.
### (D) Find the velocity of the object at any time [tex]\( t \)[/tex].
Velocity is the derivative of position with respect to time:
[tex]\[ v(t) = \frac{dx(t)}{dt} = \frac{d}{dt}(16t - 6t^2) = 16 - 12t \][/tex]
Now, find the velocity at [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex]:
[tex]\[ v(0) = 16 - 12(0) = 16 \][/tex]
[tex]\[ v(2) = 16 - 12(2) = 16 - 24 = -8 \][/tex]
So, the velocity at [tex]\( t = 0 \)[/tex] is [tex]\( 16 \)[/tex] meters per second, and the velocity at [tex]\( t = 2 \)[/tex] seconds is [tex]\( -8 \)[/tex] meters per second.
### (E) At what time and position will the body be at rest?
The body is at rest when its velocity is zero:
[tex]\[ 16 - 12t = 0 \][/tex]
[tex]\[ 12t = 16 \][/tex]
[tex]\[ t = \frac{16}{12} = \frac{4}{3} = 1.33 \][/tex]
So, the body is at rest at [tex]\( t = 1.33 \)[/tex] seconds.
To find the position at rest, substitute [tex]\( t = 1.33 \)[/tex] into the position function:
[tex]\[ x(1.33) = 16(1.33) - 6(1.33)^2 = 21.28 - 10.62 = 10.66 \][/tex]
The body is at rest at [tex]\( t = 1.33 \)[/tex] seconds and the position is [tex]\( 10.67 \)[/tex] meters.
### (F) Find the acceleration of the body at any time [tex]\( t \)[/tex].
Acceleration is the derivative of velocity with respect to time:
[tex]\[ a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(16 - 12t) = -12 \][/tex]
So, the acceleration of the body is [tex]\( -12 \)[/tex] meters per second squared at any time [tex]\( t \)[/tex].
### (G) When is the acceleration of the body zero?
The acceleration equation is [tex]\( a(t) = -12 \)[/tex], which is constant and does not depend on [tex]\( t \)[/tex].
Therefore, the acceleration is never zero.
In summary:
- (A) Position at [tex]\( t = 1 \)[/tex] s: [tex]\( 10 \)[/tex] meters
- (B) Time at the origin: [tex]\( 0 \)[/tex] seconds and [tex]\( 2.67 \)[/tex] seconds
- (C) Average velocity between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex] seconds: [tex]\( 4 \)[/tex] meters per second
- (D) Velocity function: [tex]\( 16 - 12t \)[/tex]
- Velocity at [tex]\( t = 0 \)[/tex]: [tex]\( 16 \)[/tex] meters per second
- Velocity at [tex]\( t = 2 \)[/tex]: [tex]\( -8 \)[/tex] meters per second
- (E) Time and position at rest: [tex]\( t = 1.33 \)[/tex] seconds, position [tex]\( 10.67 \)[/tex] meters
- (F) Acceleration: [tex]\( -12 \)[/tex] meters per second squared
- (G) Acceleration is never zero
### (A) Find the position of the body at [tex]\( t = 1 \)[/tex] s.
The position function is given by:
[tex]\[ x(t) = 16t - 6t^2 \][/tex]
Substitute [tex]\( t = 1 \)[/tex]:
[tex]\[ x(1) = 16(1) - 6(1)^2 \][/tex]
[tex]\[ x(1) = 16 - 6 \][/tex]
[tex]\[ x(1) = 10 \][/tex]
So, the position of the body at [tex]\( t = 1 \)[/tex] s is [tex]\( 10 \)[/tex] meters.
### (B) At what time does the body pass the origin?
To find when the body passes the origin, set [tex]\( x(t) = 0 \)[/tex]:
[tex]\[ 16t - 6t^2 = 0 \][/tex]
[tex]\[ 2t(8 - 3t) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ t = \frac{8}{3} = 2.67 \][/tex]
So, the body passes the origin at [tex]\( t = 0 \)[/tex] seconds and [tex]\( t = 2.67 \)[/tex] seconds.
### (C) Calculate the average velocity of the body between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex] seconds.
The average velocity is given by the change in position divided by the change in time:
[tex]\[ \text{Average Velocity} = \frac{x(2) - x(0)}{2 - 0} \][/tex]
First, find [tex]\( x(2) \)[/tex] and [tex]\( x(0) \)[/tex]:
[tex]\[ x(0) = 16(0) - 6(0)^2 = 0 \][/tex]
[tex]\[ x(2) = 16(2) - 6(2)^2 = 32 - 24 = 8 \][/tex]
So,
[tex]\[ \text{Average Velocity} = \frac{8 - 0}{2 - 0} = \frac{8}{2} = 4 \][/tex]
The average velocity between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex] seconds is [tex]\( 4 \)[/tex] meters per second.
### (D) Find the velocity of the object at any time [tex]\( t \)[/tex].
Velocity is the derivative of position with respect to time:
[tex]\[ v(t) = \frac{dx(t)}{dt} = \frac{d}{dt}(16t - 6t^2) = 16 - 12t \][/tex]
Now, find the velocity at [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex]:
[tex]\[ v(0) = 16 - 12(0) = 16 \][/tex]
[tex]\[ v(2) = 16 - 12(2) = 16 - 24 = -8 \][/tex]
So, the velocity at [tex]\( t = 0 \)[/tex] is [tex]\( 16 \)[/tex] meters per second, and the velocity at [tex]\( t = 2 \)[/tex] seconds is [tex]\( -8 \)[/tex] meters per second.
### (E) At what time and position will the body be at rest?
The body is at rest when its velocity is zero:
[tex]\[ 16 - 12t = 0 \][/tex]
[tex]\[ 12t = 16 \][/tex]
[tex]\[ t = \frac{16}{12} = \frac{4}{3} = 1.33 \][/tex]
So, the body is at rest at [tex]\( t = 1.33 \)[/tex] seconds.
To find the position at rest, substitute [tex]\( t = 1.33 \)[/tex] into the position function:
[tex]\[ x(1.33) = 16(1.33) - 6(1.33)^2 = 21.28 - 10.62 = 10.66 \][/tex]
The body is at rest at [tex]\( t = 1.33 \)[/tex] seconds and the position is [tex]\( 10.67 \)[/tex] meters.
### (F) Find the acceleration of the body at any time [tex]\( t \)[/tex].
Acceleration is the derivative of velocity with respect to time:
[tex]\[ a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(16 - 12t) = -12 \][/tex]
So, the acceleration of the body is [tex]\( -12 \)[/tex] meters per second squared at any time [tex]\( t \)[/tex].
### (G) When is the acceleration of the body zero?
The acceleration equation is [tex]\( a(t) = -12 \)[/tex], which is constant and does not depend on [tex]\( t \)[/tex].
Therefore, the acceleration is never zero.
In summary:
- (A) Position at [tex]\( t = 1 \)[/tex] s: [tex]\( 10 \)[/tex] meters
- (B) Time at the origin: [tex]\( 0 \)[/tex] seconds and [tex]\( 2.67 \)[/tex] seconds
- (C) Average velocity between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2 \)[/tex] seconds: [tex]\( 4 \)[/tex] meters per second
- (D) Velocity function: [tex]\( 16 - 12t \)[/tex]
- Velocity at [tex]\( t = 0 \)[/tex]: [tex]\( 16 \)[/tex] meters per second
- Velocity at [tex]\( t = 2 \)[/tex]: [tex]\( -8 \)[/tex] meters per second
- (E) Time and position at rest: [tex]\( t = 1.33 \)[/tex] seconds, position [tex]\( 10.67 \)[/tex] meters
- (F) Acceleration: [tex]\( -12 \)[/tex] meters per second squared
- (G) Acceleration is never zero
