Average power ($\overline{P}=W/(t_f-t_i)$)

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 $x(t)=a{t}^2/2$. ThusThis 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. ^[ref]