I really have NO idea what is wrong with me. I swore that I knew what I was doing but I guess not cause I keep getting the answers wrong.
1st question:
An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 8:00 A.M. will produce
Q(t) = -t3 + 8t2 + 14t
units t hours later. Compute the workers rate of production at 10:15 A.M.
First I need to find the second derivative
which is
Q(t) -3t^2+16t+14
I then subtract 10:15AM-8AM and I have a 2hr. 15min. difference: in other words t=2.15
I then plug that answer into the equation of the second derivative:
Q(2.15)=-3(2.15)^2+16(2.15)+14= 34.5325
What am I doing wrong?
2nd question:
If an object is thrown vertically, from a initial height of 165 feet
and an initial velocity of 189 feet per second, its height (in feet)
after t seconds is given by the formula
H(t) = -16t2 + 189t + 165.
(The fact that the velocity is positive indicates that the object starts
with an upward motion, if the the object is thrown downwards its
initial velocity is negative )
Find the acceleration of the object at t = 4
First of alll I know that the acceleration is shown by -16t^2 and 189t=initial speed and 165= initial height
Once again I need to get the 2nd derivative:
H(t)-32t+189
since acceleration is shown by in the 2nd derivative as -32t I just plug in 4 and that is my answer:H(4)= -32(4)=-128
What am I doing wrong here.
I really appreciate you help everyone. I hope I am not asking to many questions today. I am just going crazy trying to stay up to date in my class.
Then I thought I would plug for in this equation
1st question:
An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 8:00 A.M. will produce
Q(t) = -t3 + 8t2 + 14t
units t hours later. Compute the workers rate of production at 10:15 A.M.
First I need to find the second derivative
which is
Q(t) -3t^2+16t+14
I then subtract 10:15AM-8AM and I have a 2hr. 15min. difference: in other words t=2.15
I then plug that answer into the equation of the second derivative:
Q(2.15)=-3(2.15)^2+16(2.15)+14= 34.5325
What am I doing wrong?
2nd question:
If an object is thrown vertically, from a initial height of 165 feet
and an initial velocity of 189 feet per second, its height (in feet)
after t seconds is given by the formula
H(t) = -16t2 + 189t + 165.
(The fact that the velocity is positive indicates that the object starts
with an upward motion, if the the object is thrown downwards its
initial velocity is negative )
Find the acceleration of the object at t = 4
First of alll I know that the acceleration is shown by -16t^2 and 189t=initial speed and 165= initial height
Once again I need to get the 2nd derivative:
H(t)-32t+189
since acceleration is shown by in the 2nd derivative as -32t I just plug in 4 and that is my answer:H(4)= -32(4)=-128
What am I doing wrong here.
I really appreciate you help everyone. I hope I am not asking to many questions today. I am just going crazy trying to stay up to date in my class.
Then I thought I would plug for in this equation