Given that s=ut+at^2 and that the rate of change of distance with time = speed, show that, v=u+at OK so it is quite simple it's a matter of differentiating the first term. But why exactly?
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simply because velocity is defined as the rate of change of position with respect to time. specifically ... $\displaystyle v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}$
If you were to differentiate $\displaystyle s(t)=ut+at^2 $ you would not get $\displaystyle v(t)=u+at $ you would get $\displaystyle v(t)=u+2at $ Because $\displaystyle s(t)= ut+\frac {1} {2} at^2 $ given that the object starts at position 0.
my bad, it was a typo.
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