Answer to Question #226630 in Differential Geometry | Topology for Aqua

Question

A particle moves in space so that at time t its position is stated as x=2t+3, y=t²+3t, z=t³+2t² Find the components of its velocity and acceleration when t=1

Answer 1

Let a particle moves in space so that at time $t$ its position is stated as $x=2t+3, y=t^2+3t, z=t^3+2t^2.$

Let us find the components of its velocity:

$v_x=x'=2, v_y=y'=2t+3, v_z=z'=3t^2+4t.$

Let us find the components of its acceleration:

$a_x=x''=0, a_y=y''=2, a_z=z''=6t+4.$

When $t=1$ the components of its velocity are $v_x=2, v_y=5, v_z=7$ and the components of its acceleration are $a_x=0, a_y=2, a_z=10.$