Answer to Question #194106 in Mechanics | Relativity for Clinton Paul

Question

.There is a node, starting at (0, 0, 0), and it's moving 10 per second on the X axis, 6 per second on the Y axis, and 8 per second on the Z axis. Please answer the follow questions.

(1) What is the motion equation of this node?

(2) what is the position of this node after 5 seconds?

Answer 1

(1) Initial velocity

$v_0 = 0 \hat{i} + 0 \hat{j}+ 0 \hat{k}$

Final velocity

$v_t = 10 \hat{i} + 6 \hat{j} + 8 \hat{k}$

Take acceleration

$a_t = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}$

According to the equation of motion

$v_t = v_0t + at \\ 10 \hat{i} + 6 \hat{j} + 8 \hat{k} = (0 \hat{i} + 0 \hat{j} + 0 \hat{k}) + (a_x \hat{i} + a_y \hat{j} + a_z \hat{k})t \\ (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) = \frac{10}{t} \hat{i} + \frac{6}{t} \hat{j} + \frac{8}{ t} \hat{k}$

(2) The position of this node after 5 seconds

t = 5

According to the equation of motion

S = v_0t + \frac{1}{2}at^2

After 5 sec

$a_x \hat{i} + a_y \hat{j} + a_z \hat{k} = \frac{10}{5} \hat{i} + \frac{6}{5} \hat{j} + \frac{8}{ 5} \hat{k} \\ = 2 \hat{i} + \frac{6}{5} \hat{j} + \frac{8}{ 5} \hat{k} \\ S = (0 \hat{i} + 0 \hat{j} + 0 \hat{k}) \times 5 + \frac{1}{2} \times (2 \hat{i} + \frac{6}{5} \hat{j} + \frac{8}{ 5} \hat{k}) \times 5^2 \\ = \frac{2 \times 25}{2} \hat{i} + \frac{6 \times 25}{5 \times 2} \hat{j} + \frac{8 \times 25}{5 \times 2} \hat{k} \\ = 25 \hat{i} + 15 \hat{j} + 20 \hat{k}$

Therefore, after 5 sec the position of the node is (25, 15, 20)