The equations for any object in three dimensional motion are given as;
$x (t) = 3 t^{2} + 5$
$y (t) = - t^{2} + 3 t - 2$
$z (t) = 2 t + 1$
Calculate magnitude of acceleration, velocity and displacement vectors at time t = 0 and t = 2.

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Solution

$x (t) = 3 t^{2} + 5$

$y (t) = - t^{2} + 3 t - 2$

$z (t) = 2 t + 1$

Displacement
$x (t) at t = 0, x (0) = 5$

$x (2) = 12 + 5$

$= 17$

$y (t) at t = 0, \Rightarrow y (0) = - 2$

$---''--- t = 2, y (2) = - 4 + 6 - 2$

$= 0$

$z (t) at t = 0, \Rightarrow z (0) = 1$

$z (2) = 5$

$v_{x} = x^{1} (t) = 6 t at t = 0, v_{x} = 0 at t = 2, v_{x} = 12$

$v_{y} = y^{1} (t) = - 2 t + 3, at t = 0, v_{y} = 2, at t = 2 v_{y} = - 1$

$v_{z} = z^{1} (t) = 2 v_{z} = 2,$

$a_{x} = \frac{d V_{x}}{d t} = 6$

$a_{y} = - 2$

$a_{z} = 0$

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