Motion graphs and derivatives

In

Here ${\displaystyle s}$ is the position of the object, and ${\displaystyle t}$ is the time. Therefore, the slope of the curve gives the change in position divided by the change in time, which is the definition of the average velocity for that interval of time on the graph. If this interval is made to be infinitesimally small, such that ${\displaystyle {\Delta s}}$ becomes ${\displaystyle {ds}}$ and ${\displaystyle {\Delta t}}$ becomes ${\displaystyle {dt}}$, the result is the instantaneous velocity at time ${\displaystyle t}$, or the derivative of the position with respect to time.

A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in ${\displaystyle y}$ over change in ${\displaystyle x}$:

${\displaystyle a={\frac {\Delta y}{\Delta x}}={\frac {\Delta v}{\Delta t}}}$

where ${\displaystyle v}$ is the velocity, and ${\displaystyle t}$ is the time. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives ${\displaystyle {\begin{matrix}{\frac {dv}{dt}}\end{matrix}}}$, the instantaneous acceleration at time ${\displaystyle t}$, or the derivative of the velocity with respect to time (or the

(${\displaystyle \mathrm {m/s^{2}} }$, usually termed "meters per second-squared").

Since the velocity of the object is the

displacement

of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the time cancels out, and only displacement remains.)

The same multiplication rule holds true for acceleration vs. time graphs. When acceleration is multiplied

Variable rates of change

Since acceleration differentiates the expression involving position, it can be rewritten as a second derivative with respect to time:

${\displaystyle a={\frac {d^{2}s}{dt^{2}}}.}$

Since, for the purposes of mechanics such as this,

${\displaystyle s(t_{2})-s(t_{1})=\int _{t_{1}}^{t_{2}}{v}\,dt,}$

${\displaystyle v(t_{2})-v(t_{1})=\int _{t_{1}}^{t_{2}}{a}\,dt.}$

See also

• Displacement (vector)
• Velocity
• Acceleration
• Kinematics

References

• Wolfson, Richard; Jay M. Pasachoff (1999). Physics for Scientists and Engineers (3rd ed.).
  ISBN 0-321-03571-2
  .