Starting from the origin at time t = 0 , with initial velocity `5hatj "ms"^(-1)` , a particle moves in the x - y plane with a constant acceleration of `(10hati+4hatj)ms^(-2)` . At time t , its coordinates are `(20 m,y_0 m)` . The value of t and `y_0` are , respectively :

Options

A. 2 s and 18 m
B. 5 s and 25 m
C. 2 s and 24 m
D. 4 s and 52 m

Answer

To solve the problem step-by-step, we will use the equations of motion to find the time \( t \) and the coordinate \( y_0 \) of the particle.

### Step 1: Identify the initial conditions and parameters
- The initial position of the particle is at the origin: \( (0, 0) \).
- The initial velocity \( \vec{u} = 5 \hat{j} \, \text{m/s} \) implies \( u_x = 0 \, \text{m/s} \) and \( u_y = 5 \, \text{m/s} \).
- The acceleration \( \vec{a} = (10 \hat{i} + 4 \hat{j}) \, \text{m/s}^2 \) gives \( a_x = 10 \, \text{m/s}^2 \) and \( a_y = 4 \, \text{m/s}^2 \).

### Step 2: Use the x-direction motion equation to find time \( t \)
The equation of motion in the x-direction is given by:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Since \( u_x = 0 \), this simplifies to:
\[ x = \frac{1}{2} a_x t^2 \]
Given that at time \( t \), the x-coordinate is \( 20 \, \text{m} \):
\[ 20 = \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 20 = 5 t^2 \]
Dividing both sides by 5:
\[ t^2 = 4 \]
Taking the square root:
\[ t = 2 \, \text{s} \]

### Step 3: Use the y-direction motion equation to find \( y_0 \)
Now, we will use the equation of motion in the y-direction:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ y = 5 \cdot 2 + \frac{1}{2} \cdot 4 \cdot (2^2) \]
Calculating each term:
\[ y = 10 + \frac{1}{2} \cdot 4 \cdot 4 \]
\[ y = 10 + \frac{16}{2} \]
\[ y = 10 + 8 \]
\[ y = 18 \, \text{m} \]

### Conclusion
The values of \( t \) and \( y_0 \) are:
\[ t = 2 \, \text{s}, \quad y_0 = 18 \, \text{m} \]

### Final Answer
The value of \( t \) and \( y_0 \) are, respectively: \( 2 \, \text{s} \) and \( 18 \, \text{m} \).

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