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