Particle Kinematics for Straight Line Motion

There are two extreme cases for particle kinematics: that for purely circular/radial motion, and that for purely straight line motion. Other cases fall between the two. This is the latter; an example of the former is here. The example comes from Theoretical Mechanics – A Short Course -Targ; the solution method, however, is different. Our goal is to find the position, velocity and acceleration vectors, and to determine their directions.

We have a particle whose motion can be defined by the time-dependent position vector

$r(t)=\left (8\,t-4\,{t}^{2}\right )i+\left (6\,t-3\,{t}^{2}\right )j$ (1)

If we consider the i and j terms as x and y equations of motion (which we can) the line of motion can be shown to be

$y = \frac{3}{4}x$ (2)

This is shown in the graph at the top of the page. The unit position vector can be obtained by dividing the right hand side of Equation (1) by the scalar magnitude of the position vector, or

$u(t)=1/5\,{\frac {\left (8\,t-4\,{t}^{2}\right )i+\left (6\,t-3\,{t}^{2}\right )j}{t\left (-2+t\right )}}$ (3)

which simplifies to

$u(t)=-4/5\,i-3/5\,j$ (4)

The velocity vector is the derivative of Equation (1), or

${\frac {d}{dt}}r(t)=\left (8-8\,t\right )i+\left (6-6\,t\right )j$ (5)

The scalar magnitude of the velocity vector is

$v(t) = -10+10\,t$ (6)

This can also be obtained by computing the dot product of the unit position vector (Equation (3)) with the velocity vector (Equation (5).) What this indicates is that the velocity vector has the same direction as the position vector, which we would expect with straight-line motion, even if it reverses as it does here.

If we derive Equation (5) we obtain the acceleration vector,

${\frac {d^{2}}{d{t}^{2}}}r(t)=-8\,i-6\,j$ (7)