In April 2025
Answer to Question #194106 in Mechanics | Relativity for Clinton Paul
.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?
(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 \\\\\n\n10 \\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 \\\\\n\n(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} \\\\\n\n= 2 \\hat{i} + \\frac{6}{5} \\hat{j} + \\frac{8}{ 5} \\hat{k} \\\\\n\nS = (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 \\\\\n\n= \\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} \\\\\n\n= 25 \\hat{i} + 15 \\hat{j} + 20 \\hat{k}"
Therefore, after 5 sec the position of the node is (25, 15, 20)
Comments
Leave a comment