Planar motion: displacement, distance, and changing speed

Understand the idea

Planar motion combines position, velocity, and acceleration with scalar measurements of travel. Componentwise derivatives describe motion along each axis. Speed and total distance depend on the magnitude of velocity, so a particle can move quickly even when its net displacement is small.

When velocity is nonzero, the derivative of speed has the sign of the dot product of velocity and acceleration. A positive dot product means speed is increasing; a negative one means it is decreasing. Checking just one acceleration component is not sufficient in the plane.

For total distance, integrate speed over the time interval. For displacement, subtract positions or integrate velocity componentwise. A particle can return to its initial horizontal coordinate while still moving vertically and traveling a nonzero distance.

v=\sqrt{(x')^2+(y')^2},\qquad \frac{dv}{dt}=\frac{\mathbf v\cdot\mathbf a}{|\mathbf v|}\quad(|\mathbf v|>0),\qquad D=\int_a^b|\mathbf v(t)|\,dt

A method you can apply

Compute both velocity components and both acceleration components.
Use velocity magnitude for speed and its dot product with acceleration to assess changing speed.
Use the requested vector displacement or scalar distance integral, with the correct time bounds.

Worked example

A particle has $\displaystyle x=t^2-2t$ and $\displaystyle y=3t$ . Find its speed and whether speed is increasing at $\displaystyle t=2$ , and set up its distance over $\displaystyle 0\le t\le2$ .

Velocity is $\displaystyle \langle2t-2,3\rangle$ and acceleration is $\displaystyle \langle2,0\rangle$ .
At $\displaystyle t=2$ , speed is $\displaystyle \sqrt{2^2+3^2}=\sqrt{13}$ .
The dot product at $\displaystyle 2$ is $\displaystyle \langle2,3\rangle\cdot\langle2,0\rangle=4>0$ , so speed is increasing.
Total distance is $\displaystyle \int_0^2\sqrt{(2t-2)^2+9}\,dt$ , while displacement is $\displaystyle \langle0,6\rangle$ .

The speed is

\displaystyle \sqrt{13}

and is increasing at

\displaystyle t=2

. Distance requires integrating the nonnegative speed, not taking only the magnitude of displacement.

Try it yourself

At an instant, $\displaystyle \mathbf v=\langle3,-4\rangle$ and $\displaystyle \mathbf a=\langle-2,1\rangle$ . Find speed and its rate of change.

Show solution and reasoning

Speed is $\displaystyle 5$ . The dot product is $\displaystyle -6-4=-10$ , so the rate of change of speed is $\displaystyle -10/5=-2$ . The particle is slowing at that instant.