The horizontal and vertical displacements of a particle moving along a curved line are given by `x=5t and y = 2t^2 + t.` Time after which its velocity vector makes an angle of `45^@` with the horizontal is

Options

A. `0.5 s`
B. `1 s`
C. `2 s`
D. `1.5 s`

Explanation

To solve the problem, we need to find the time \( t \) when the velocity vector of the particle makes an angle of \( 45^\circ \) with the horizontal.

### Step-by-Step Solution:

1. **Identify the Displacement Equations**:
The horizontal and vertical displacements are given by:
\[ x = 5t \]
\[ y = 2t^2 + t \]

2. **Find the Velocity Components**:
The velocity components can be found by differentiating the displacement equations with respect to time \( t \).
- The horizontal component of velocity \( V_x \) is:
\[ V_x = \frac{dx}{dt} = \frac{d(5t)}{dt} = 5 \]
- The vertical component of velocity \( V_y \) is:
\[ V_y = \frac{dy}{dt} = \frac{d(2t^2 + t)}{dt} = 4t + 1 \]

3. **Condition for \( 45^\circ \) Angle**:
For the velocity vector to make an angle of \( 45^\circ \) with the horizontal, the tangent of the angle must equal 1:
\[ \tan(45^\circ) = 1 = \frac{V_y}{V_x} \]
This implies:
\[ V_y = V_x \]

4. **Set Up the Equation**:
Substitute the expressions for \( V_x \) and \( V_y \):
\[ 4t + 1 = 5 \]

5. **Solve for \( t \)**:
Rearranging the equation:
\[ 4t = 5 - 1 \]
\[ 4t = 4 \]
\[ t = 1 \text{ second} \]

6. **Conclusion**:
The time after which the velocity vector makes an angle of \( 45^\circ \) with the horizontal is \( t = 1 \) second.