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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]

Instantaneous power is power at a particular moment. Consider a point mass with mass $m=2\mathrm{kg}$, initially at rest, acted on by a constant net force $F=3\mathrm{N}$ along a straight path. Newton's second law gives constant acceleration $a=F/m=1.5m/{\mathrm{s}}^2$, hence velocity $v(t)=at$ andThe instantaneous power starts at zero even though the force is nonzero, because the initial velocity is zero. At distance $4\mathrm{m}$, the kinetic energy example below gives speed $\sqrt{12}m/s$, so the final instantaneous power is $3\sqrt{12}\mathrm{W}\approx 10.3923\mathrm{W}$. Negative instantaneous power means the selected force is removing energy at that moment.

Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]