A particle leaves the origin with initial velocity `vec(v)_(0)=11hat(i)+14 hat(j) m//s`. It undergoes a constant acceleration given by `vec(a)=-22/5 hat(i)+2/15 hat(j) m//s^(2)`.
When does the particle cross the y-axis ?

Options

A. 2 sec
B. 4 sec
C. 5 sec
D. 7 sec

Answer

To solve the problem of when the particle crosses the y-axis, we need to analyze the motion of the particle in the x and y directions separately. The particle starts at the origin with an initial velocity and experiences a constant acceleration.

### Step-by-step Solution:

1. **Identify the Initial Conditions**:
- Initial velocity: \( \vec{v}_0 = 11 \hat{i} + 14 \hat{j} \, \text{m/s} \)
- Initial position: \( \vec{r}_0 = 0 \hat{i} + 0 \hat{j} \, \text{m} \)
- Acceleration: \( \vec{a} = -\frac{22}{5} \hat{i} + \frac{2}{15} \hat{j} \, \text{m/s}^2 \)

2. **Determine the Displacement Equations**:
The displacement in the x-direction can be described by the equation of motion:
\[ x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \]
Where:
- \( x_0 = 0 \)
- \( v_{0x} = 11 \, \text{m/s} \)
- \( a_x = -\frac{22}{5} \, \text{m/s}^2 \)

Thus, the equation becomes:
\[ x(t) = 11t - \frac{22}{10} t^2 = 11t - \frac{11}{5} t^2 \]

3. **Set the x-displacement to Zero**:
To find when the particle crosses the y-axis, we set \( x(t) = 0 \):
\[ 11t - \frac{11}{5} t^2 = 0 \]

4. **Factor the Equation**:
Factoring out \( t \):
\[ t \left( 11 - \frac{11}{5} t \right) = 0 \]
This gives us two solutions:
- \( t = 0 \) (the initial time)
- \( 11 - \frac{11}{5} t = 0 \)

5. **Solve for t**:
Solving the second equation:
\[ 11 = \frac{11}{5} t \]
\[ t = 11 \times \frac{5}{11} = 5 \, \text{seconds} \]

6. **Conclusion**:
The particle crosses the y-axis at \( t = 5 \) seconds.

### Final Answer:
The particle crosses the y-axis at \( t = 5 \) seconds.

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