= Average power
{title2=$\overline P=W/(t_f-t_i)$}
<Average power> divides the <work> over a nonzero interval by the elapsed <time>. For the same $2\,\mathrm{kg}$ <point mass> driven from rest through $4\,\mathrm m$ by a net $3\,\mathrm N$ <force>, the <displacement> is $x(t)=at^2/2$. Thus
$$
t_f=\sqrt{\frac{2(4\,\mathrm m)}{1.5\,\mathrm{m/s^2}}}
=\sqrt{16/3}\,\mathrm s\approx2.309401\,\mathrm s,
\qquad \overline P=\frac{12\,\mathrm J}{t_f}
\approx5.19615\,\mathrm W.
$$
This <average power> is half the final <instantaneous power> because $P(t)$ is a <linear function> of <time>, starting at zero. Its <graph of a function> bounds a <triangle> of area $P(t_f)t_f/2=12\,\mathrm J$; the horizontal line at <average power> encloses a <rectangle> with the same area. In general, an endpoint value of <instantaneous power> does not determine <average power>. https://openstax.org/books/university-physics-volume-1/pages/7-4-power[OpenStax University Physics 1, §7.4: Average and instantaneous power]{ref}
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