Position, velocity and acceleration

Recall that when $${\bf r}(t) \ = \ x(t)\,{\bf i} + y(t)\,{\bf j} + z(t)\,{\bf k}$$ is a vector function whose values are vectors in ${\mathbb R}^3$, then a space curve is simply the path traced out by the tip of ${\bf r}(t)$ as $t$ varies. If this path is thought of as the trajectory of an object moving in space, the parameter $t$ is naturally time, so that the tip of ${\bf r}(t)$ gives the position of the object at time $t$. In terms of motion in space, the examples that we saw earlier,

${\bf r}(t) = \langle \cos t,\, \sin t, t \rangle$

${\bf r}(t) = \langle t \cos t,\, t \sin t, t \rangle$

Example 1: Find the position ${\bf r}(t)$ of a particle with acceleration $${\bf a}(t) \,=\, 2\, {\bf i} + 12 t\, {\bf j}\,,$$ when its initial velocity and position satisfy $${\bf v} (0)\,=\, 7\, {\bf i}, \quad {\bf r}(0)\,=\, 2\, {\bf i} + 9 {\bf k}\,.$$ Solution: Since ${\bf a}(t) = {\bf v}'(t)$, integration gives $${\bf v}(t)\, = {\bf v}(0) + \int_0^t {\bf a}(s) ds \, = \,2t\, {\bf i}+ 6t^2\, {\bf j} +{\bf v}(0)\, =\, 2t\, {\bf i} + 6t^2\, {\bf j} + 7\, {\bf i}\,.$$

Likewise, we integrate the velocity to get the position: $${\bf r}(t) \ = \ \int\, {\bf v}(t)\, dt \ = \ t^2\, {\bf i} + 2t^3\, {\bf j} + 7t\, {\bf i} + {\bf r}(0)\,.$$ Consequently, $${\bf r}(t) \ = \ t^2\, {\bf i} + 2t^3\, {\bf j} + 7t\, {\bf i} + (2\, {\bf i} + 9 {\bf k})$$ $$\qquad = \ (t^2 +7t +2)\, {\bf i} + 2t^3\, {\bf j} + 9\, {\bf k}\,,$$ or $$x(t) = t^2+7t+2; \qquad y(t) = 2t^3; \qquad z(t) = 9\,.$$