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OpenStudy (anonymous):
acceleration = 3t^2 + 2t + 1
find expression for instantaneous velocity at any time "t"
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OpenStudy (anonymous):
I got the answer but here is integration constant included
OpenStudy (anonymous):
possibly. maybe there's more to this question than you're telling us :)
OpenStudy (anonymous):
no nothing more, i just wanted to know if integration constatnt should be there or not
OpenStudy (anonymous):
that's a stumper.
Maybe just set it to zero?
OpenStudy (anonymous):
In the options it is given with and without a constant
and the correct answer is the option without a constant
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OpenStudy (anonymous):
set that to zero for what
OpenStudy (anonymous):
i don't want to find max or min
OpenStudy (anonymous):
i want to know then why integration constant is not included
OpenStudy (anonymous):
I meant set the constant to zero... thereby having pretty much no constant lawl
OpenStudy (anonymous):
why constant is 0
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OpenStudy (anonymous):
I don't know... I'm just as confused as you are :D
OpenStudy (anonymous):
In Newtonian mechanics we don't assume constant
OpenStudy (anonymous):
Like in f=dp/dt
f=kma when mass is constant
k is 1 Newton
so f= ma
OpenStudy (anonymous):
I don't know the reason
ganeshie8 (ganeshie8):
\[\large v(t) - v(0) = \int \limits_0^t 3t^2 + 2t + 1 ~dt\]
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OpenStudy (anonymous):
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