A particle at origin (0,0) moving with initial velocity `u = 5 m/s hatj` and acceleration `10 hati + 4 hatj`. After t time it reaches at position (20,y) then find t & y

Options

A. t = 2, y = 18
B. t = 4, y = 16
C. t = 6, y = 12
D. t =8, y =10

Answer

To solve the problem step by step, we will use the equations of motion for both the x and y coordinates.

### Step 1: Identify the initial conditions and parameters
- Initial velocity \( \mathbf{u} = 5 \, \hat{j} \, \text{m/s} \) (which means \( u_x = 0 \, \text{m/s} \) and \( u_y = 5 \, \text{m/s} \))
- Acceleration \( \mathbf{a} = 10 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2 \) (which means \( a_x = 10 \, \text{m/s}^2 \) and \( a_y = 4 \, \text{m/s}^2 \))
- Final position after time \( t \) is \( (20, y) \)

### Step 2: Use the equation of motion for the x-coordinate
The equation of motion in the x-direction is given by:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Since \( u_x = 0 \):
\[ x = \frac{1}{2} a_x t^2 \]
Substituting \( x = 20 \) and \( a_x = 10 \):
\[ 20 = \frac{1}{2} \cdot 10 \cdot t^2 \]
\[ 20 = 5 t^2 \]
\[ t^2 = \frac{20}{5} = 4 \]
\[ t = 2 \, \text{s} \]

### Step 3: Use the equation of motion for the y-coordinate
Now we will find the y-coordinate using the equation of motion in the y-direction:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting \( u_y = 5 \, \text{m/s} \), \( a_y = 4 \, \text{m/s}^2 \), and \( t = 2 \):
\[ y = 5 \cdot 2 + \frac{1}{2} \cdot 4 \cdot (2)^2 \]
Calculating:
\[ y = 10 + \frac{1}{2} \cdot 4 \cdot 4 \]
\[ y = 10 + \frac{1}{2} \cdot 16 \]
\[ y = 10 + 8 = 18 \]

### Final Result
Thus, the values of \( t \) and \( y \) are:
\[ t = 2 \, \text{s}, \quad y = 18 \]

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