- #1
lxman
- 77
- 0
Hello all,
I am a first year college student who has just been through Calc I, and I am working to put some concepts together. I am also new to the forum. I am working out a thought experiment, and I have what I believe is a correct solution. I would appreciate it if someone could verify my calculations.
Given a 1kg mass with an acceleration of sin(t), we can then derive that its velocity would be -cos(t) and its position function would be -sin(t). So we have the following:
a(t) = sin(t)
v(t) = -cos(t)
s(t) = -sin(t)
Now, for my purposes, as I will be calculating power used to move this object around, I don't want a negative position, therefore, since the function s is an integral of the function v, I will add a constant of one. Thus I now have:
a(t) = sin(t)
v(t) = -cos(t)
s(t) = -sin(t) + 1
Kinetic energy is 1/2*m*v^2. Therefore I obtain:
J(t) = 1/2 * (-cos(t)) ^ 2
Work is energy expended over time. Therefore I integrate again, and obtain:
W(t) = 1/4 * t + 1/8 * sin(2 * t)
If I understand correctly, I have now obtained how many watts are being expended at time t. The electric company, for instance, does not bill by kW. They bill by kWh. So in order to get the total wattage over time I would need to integrate once again to arrive at:
Ws(t) = 1/8 * t ^ 2 - 1/16 * cos(2 * t)
This, then, should give me the total work done at time t to oscillate my 1kg mass back and forth. Am I deriving this correctly?
I am a first year college student who has just been through Calc I, and I am working to put some concepts together. I am also new to the forum. I am working out a thought experiment, and I have what I believe is a correct solution. I would appreciate it if someone could verify my calculations.
Given a 1kg mass with an acceleration of sin(t), we can then derive that its velocity would be -cos(t) and its position function would be -sin(t). So we have the following:
a(t) = sin(t)
v(t) = -cos(t)
s(t) = -sin(t)
Now, for my purposes, as I will be calculating power used to move this object around, I don't want a negative position, therefore, since the function s is an integral of the function v, I will add a constant of one. Thus I now have:
a(t) = sin(t)
v(t) = -cos(t)
s(t) = -sin(t) + 1
Kinetic energy is 1/2*m*v^2. Therefore I obtain:
J(t) = 1/2 * (-cos(t)) ^ 2
Work is energy expended over time. Therefore I integrate again, and obtain:
W(t) = 1/4 * t + 1/8 * sin(2 * t)
If I understand correctly, I have now obtained how many watts are being expended at time t. The electric company, for instance, does not bill by kW. They bill by kWh. So in order to get the total wattage over time I would need to integrate once again to arrive at:
Ws(t) = 1/8 * t ^ 2 - 1/16 * cos(2 * t)
This, then, should give me the total work done at time t to oscillate my 1kg mass back and forth. Am I deriving this correctly?