Answer :
The position vector of the particle is given by:
[tex]\[ r(t) = 3t \hat{i} + 2t^2 \hat{j} + 5 \hat{k} \][/tex]
To find the velocity [tex]\( v(t) \)[/tex], we need to take the derivative of the position vector with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}\left(3t \hat{i} + 2t^2 \hat{j} + 5 \hat{k}\right) \][/tex]
Computing the derivative term by term, we get:
[tex]\[ v(t) = \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(2t^2) \hat{j} + \frac{d}{dt}(5) \hat{k} \][/tex]
[tex]\[ v(t) = 3 \hat{i} + 4t \hat{j} + 0 \hat{k} \][/tex]
[tex]\[ v(t) = 3 \hat{i} + 4t \hat{j} \][/tex]
Now, we want to find [tex]\( v(t) \)[/tex] at [tex]\( t = 1 \)[/tex] second:
[tex]\[ v(1) = 3 \hat{i} + 4 \cdot 1 \hat{j} \][/tex]
[tex]\[ v(1) = 3 \hat{i} + 4 \hat{j} \][/tex]
The components of the velocity vector at [tex]\( t = 1 \)[/tex] second are:
[tex]\[ v_x = 3, \quad v_y = 4, \quad v_z = 0 \][/tex]
Next, we need to find the direction angles of this velocity vector. Let's start with the angle with the [tex]\( x \)[/tex]-axis:
[tex]\[ \theta_x = \tan^{-1}\left(\frac{|v_y|}{|v_x|}\right) \][/tex]
[tex]\[ \theta_x = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \][/tex]
Then, we can find the angle with the [tex]\( y \)[/tex]-axis:
[tex]\[ \theta_y = \tan^{-1}\left(\frac{|v_x|}{|v_y|}\right) \][/tex]
[tex]\[ \theta_y = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ \][/tex]
Now, let's compare these angles with the given choices:
1. [tex]\( 45^\circ \)[/tex] with [tex]\( X \)[/tex]-axis
2. [tex]\( 63^\circ \)[/tex] with [tex]\( Y \)[/tex]-axis
3. [tex]\( 30^\circ \)[/tex] with [tex]\( Y \)[/tex]-axis
4. [tex]\( 53^\circ \)[/tex] with [tex]\( X \)[/tex]-axis
We find that the direction of [tex]\( v(1) \)[/tex] corresponds to [tex]\( 53^\circ \)[/tex] with the [tex]\( X \)[/tex]-axis. Therefore, the correct choice is:
4) [tex]\( 53^\circ \)[/tex] with [tex]\( X \)[/tex]-axis
[tex]\[ r(t) = 3t \hat{i} + 2t^2 \hat{j} + 5 \hat{k} \][/tex]
To find the velocity [tex]\( v(t) \)[/tex], we need to take the derivative of the position vector with respect to time [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}\left(3t \hat{i} + 2t^2 \hat{j} + 5 \hat{k}\right) \][/tex]
Computing the derivative term by term, we get:
[tex]\[ v(t) = \frac{d}{dt}(3t) \hat{i} + \frac{d}{dt}(2t^2) \hat{j} + \frac{d}{dt}(5) \hat{k} \][/tex]
[tex]\[ v(t) = 3 \hat{i} + 4t \hat{j} + 0 \hat{k} \][/tex]
[tex]\[ v(t) = 3 \hat{i} + 4t \hat{j} \][/tex]
Now, we want to find [tex]\( v(t) \)[/tex] at [tex]\( t = 1 \)[/tex] second:
[tex]\[ v(1) = 3 \hat{i} + 4 \cdot 1 \hat{j} \][/tex]
[tex]\[ v(1) = 3 \hat{i} + 4 \hat{j} \][/tex]
The components of the velocity vector at [tex]\( t = 1 \)[/tex] second are:
[tex]\[ v_x = 3, \quad v_y = 4, \quad v_z = 0 \][/tex]
Next, we need to find the direction angles of this velocity vector. Let's start with the angle with the [tex]\( x \)[/tex]-axis:
[tex]\[ \theta_x = \tan^{-1}\left(\frac{|v_y|}{|v_x|}\right) \][/tex]
[tex]\[ \theta_x = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \][/tex]
Then, we can find the angle with the [tex]\( y \)[/tex]-axis:
[tex]\[ \theta_y = \tan^{-1}\left(\frac{|v_x|}{|v_y|}\right) \][/tex]
[tex]\[ \theta_y = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ \][/tex]
Now, let's compare these angles with the given choices:
1. [tex]\( 45^\circ \)[/tex] with [tex]\( X \)[/tex]-axis
2. [tex]\( 63^\circ \)[/tex] with [tex]\( Y \)[/tex]-axis
3. [tex]\( 30^\circ \)[/tex] with [tex]\( Y \)[/tex]-axis
4. [tex]\( 53^\circ \)[/tex] with [tex]\( X \)[/tex]-axis
We find that the direction of [tex]\( v(1) \)[/tex] corresponds to [tex]\( 53^\circ \)[/tex] with the [tex]\( X \)[/tex]-axis. Therefore, the correct choice is:
4) [tex]\( 53^\circ \)[/tex] with [tex]\( X \)[/tex]-axis
