At time t = 0 s, a disk is sliding on horizontal table with a velocity 3.00 m/s, `65.0^(@)` above the +x axis. As the disk slides, a constant acceleration acts on it that has the following components: `a_(x)=-0.460m//s^(2)anda_(y)=-0.980m//s^(2)`. What is the velocity of the puck at time t = 1.50 s?

Options

A. 1.83 m/s, `62.0^(@)` above the +x axis
B. 2.04 m/s, `71.3^(@)` above the +x axis
C. 1.38 m/s, `65.2^(@)` above the +x axis
D. 1.06 m/s, `58.7^(@)` above the +x axis

Answer

To find the velocity of the disk at time \( t = 1.50 \, \text{s} \), we will follow these steps:

### Step 1: Determine the initial velocity components
The initial velocity \( \vec{u} \) is given as \( 3.00 \, \text{m/s} \) at an angle of \( 65^\circ \) above the positive x-axis. We can break this velocity into its x and y components using trigonometric functions:

\[ u_x = u \cos(\theta) = 3.00 \cos(65^\circ) \]
\[ u_y = u \sin(\theta) = 3.00 \sin(65^\circ) \]

Calculating these values:
\[ u_x = 3.00 \cos(65^\circ) \approx 3.00 \times 0.4226 \approx 1.27 \, \text{m/s} \]
\[ u_y = 3.00 \sin(65^\circ) \approx 3.00 \times 0.9063 \approx 2.72 \, \text{m/s} \]

### Step 2: Identify the acceleration components
The acceleration components are given as:
\[ a_x = -0.460 \, \text{m/s}^2 \]
\[ a_y = -0.980 \, \text{m/s}^2 \]

### Step 3: Calculate the final velocity components at \( t = 1.50 \, \text{s} \)
Using the first equation of motion:
\[ v_x = u_x + a_x \cdot t \]
\[ v_y = u_y + a_y \cdot t \]

Substituting the values:
\[ v_x = 1.27 + (-0.460) \cdot (1.50) = 1.27 - 0.69 \approx 0.58 \, \text{m/s} \]
\[ v_y = 2.72 + (-0.980) \cdot (1.50) = 2.72 - 1.47 \approx 1.25 \, \text{m/s} \]

### Step 4: Calculate the magnitude of the final velocity
The magnitude of the final velocity \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(0.58)^2 + (1.25)^2} = \sqrt{0.3364 + 1.5625} = \sqrt{1.8989} \approx 1.38 \, \text{m/s} \]

### Step 5: Calculate the angle of the final velocity
The angle \( \phi \) with respect to the positive x-axis can be calculated using the tangent function:
\[ \phi = \tan^{-1}\left(\frac{v_y}{v_x}\right) \]
Substituting the values:
\[ \phi = \tan^{-1}\left(\frac{1.25}{0.58}\right) \approx \tan^{-1}(2.16) \approx 65.0^\circ \]

### Final Result
Thus, the velocity of the disk at time \( t = 1.50 \, \text{s} \) is approximately:
\[ \text{Velocity} \approx 1.38 \, \text{m/s} \, \text{at} \, 65.0^\circ \, \text{above the positive x-axis} \]