Donald G. Shead - most prolific member of Newsgroup.

"ghytrfvbnmju7654" wrote in message
m...

Interesting handle, there - typed entirely with the index fingers of each
hand, in a counterclockwise spiral...
16 characters and, once you have typed it, impossible to forget.


Tom Davidson
Richmond, VA

"Donald G. Shead" wrote in message om...
"ghytrfvbnmju7654" wrote in message
m...
Remember, in any case, that the formula only works for constant
acceleration. You can't start accelerating something at
2000 m/s^2 for a while and then switch to 2 m/s^2, and then
plug that into the formula as if the object had been
accelerating like that all along.

Don't be asinine! That's the kind of thing some people think can be done
with the calculus.

If you mean that the problem can be solved with calculus, that
is true. Of course, calculus does not commit that mathematical
fallacy when solving the problem. It takes into account the
value of the acceleration at every moment.

Not everyone needs to learn calculus, but anyone who needs to be
able to solve this type of problem does.

"ghytrfvbnmju7654" wrote in message
m...
"Donald G. Shead" wrote in message
om...
"ghytrfvbnmju7654" wrote in message
m...
Remember, in any case, that the formula only works for constant
acceleration. You can't start accelerating something at
2000 m/s^2 for a while and then switch to 2 m/s^2, and then
plug that into the formula as if the object had been
accelerating like that all along.

Don't be asinine! That's the kind of thing some people think can be done
with the calculus.

If you mean that the problem can be solved with calculus, that
is true. Of course, calculus does not commit that mathematical
fallacy when solving the problem. It takes into account the
value of the acceleration at every moment.

Not everyone needs to learn calculus, but anyone who needs to be
able to solve this type of problem does.

Oh pooh! A problem of that type only has to be separated into two parts: The
2000 m/s^2 for a while part, and the 2 m/s^2 part. That's not calculus; is
it(;^?

"Donald G. Shead" wrote in message om...
Oh pooh! A problem of that type only has to be separated into two parts: The
2000 m/s^2 for a while part, and the 2 m/s^2 part. That's not calculus; is
it(;^?

Yes, this would work, if the force changed instantaneously, but you'd
have to put a term for initial velocity into your formula.
It wouldn't work if the acceleration was changing slowly. Let the
acceleration be 2000 m/s^2 - (2 m/s^3) * (the time elapsed).
Where is the object when the acceleration is 2 m/s^2 ?