A particle is moving in x-y plane. Its initial velocity and acceleration are `u=(4 hati+8 hatj) m//s and a=(2 hati-4 hatj) m//s^2.` Find
(a) the time when the particle will cross the x-axis.
(b) x-coordinate of particle at this instant.
(c) velocity of the particle at this instant.
Initial coordinates of particle are `(4m,10m).`

Explanation

To solve the problem step by step, we will break it down into three parts as specified in the question.

### Given Data:
- Initial velocity, **u** = \(4 \hat{i} + 8 \hat{j}\) m/s
- Acceleration, **a** = \(2 \hat{i} - 4 \hat{j}\) m/s²
- Initial coordinates of the particle = (4 m, 10 m)

### (a) Time when the particle will cross the x-axis

To find the time when the particle crosses the x-axis, we need to determine when the y-coordinate becomes zero. The equation of motion in the y-direction is given by:

\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]

Where:
- \(s_y\) = displacement in the y-direction
- \(u_y\) = initial velocity in the y-direction = 8 m/s
- \(a_y\) = acceleration in the y-direction = -4 m/s²

The initial y-coordinate is 10 m, and we want to find the time when the particle crosses the x-axis (y = 0). Thus, the displacement \(s_y\) will be:

\[ s_y = 0 - 10 = -10 \text{ m} \]

Substituting the values into the equation:

\[ -10 = 8t + \frac{1}{2}(-4)t^2 \]

This simplifies to:

\[ -10 = 8t - 2t^2 \]

Rearranging gives:

\[ 2t^2 - 8t - 10 = 0 \]

Dividing the entire equation by 2:

\[ t^2 - 4t - 5 = 0 \]

Now, we can factor the quadratic equation:

\[ (t - 5)(t + 1) = 0 \]

This gives us two solutions:

\[ t = 5 \quad \text{and} \quad t = -1 \]

Since time cannot be negative, we have:

\[ t = 5 \text{ seconds} \]

### (b) x-coordinate of the particle at this instant

To find the x-coordinate when \(t = 5\) seconds, we use the equation of motion in the x-direction:

\[ s_x = u_x t + \frac{1}{2} a_x t^2 \]

Where:
- \(u_x\) = initial velocity in the x-direction = 4 m/s
- \(a_x\) = acceleration in the x-direction = 2 m/s²

The initial x-coordinate is 4 m. Thus,

\[ s_x = x_i + u_x t + \frac{1}{2} a_x t^2 \]

Substituting the values:

\[ x = 4 + 4(5) + \frac{1}{2}(2)(5^2) \]

Calculating:

\[ x = 4 + 20 + \frac{1}{2}(2)(25) \]
\[ x = 4 + 20 + 25 \]
\[ x = 49 \text{ m} \]

### (c) Velocity of the particle at this instant

The velocity of the particle at time \(t\) is given by:

\[ v = u + at \]

Substituting the values:

\[ v = (4 \hat{i} + 8 \hat{j}) + (2 \hat{i} - 4 \hat{j}) \cdot 5 \]

Calculating:

\[ v = (4 + 10) \hat{i} + (8 - 20) \hat{j} \]
\[ v = 14 \hat{i} - 12 \hat{j} \text{ m/s} \]

### Summary of Results:
- (a) Time when the particle crosses the x-axis: **5 seconds**
- (b) x-coordinate at this instant: **49 m**
- (c) Velocity at this instant: **14 i - 12 j m/s**