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 analyze the motion of the particle in the x and y directions separately.

### Step 1: Analyze the motion in the x-direction
The particle starts from the origin (0,0) with an initial velocity of \(0 \, \text{m/s}\) in the x-direction and has a constant acceleration of \(10 \, \text{m/s}^2\).

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(t\) is the time.

For the x-direction:
- \(s = 20 \, \text{m}\)
- \(u = 0 \, \text{m/s}\)
- \(a = 10 \, \text{m/s}^2\)

Substituting these values into the equation:
\[ 20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 20 = 5t^2 \]
Dividing both sides by 5:
\[ t^2 = 4 \]
Taking the square root:
\[ t = 2 \, \text{s} \]

### Step 2: Analyze the motion in the y-direction
Now, we will find the y-coordinate \(y_0\) at \(t = 2 \, \text{s}\). The initial velocity in the y-direction is \(5 \, \text{m/s}\) and the acceleration is \(4 \, \text{m/s}^2\).

Using the same equation of motion:
\[ y_0 = ut + \frac{1}{2} a t^2 \]
For the y-direction:
- \(u = 5 \, \text{m/s}\)
- \(a = 4 \, \text{m/s}^2\)
- \(t = 2 \, \text{s}\)

Substituting these values into the equation:
\[ y_0 = 5 \cdot 2 + \frac{1}{2} \cdot 4 \cdot (2)^2 \]
Calculating each term:
\[ y_0 = 10 + \frac{1}{2} \cdot 4 \cdot 4 \]
\[ y_0 = 10 + 8 \]
\[ y_0 = 18 \, \text{m} \]

### Final Result
Thus, the values of \(t\) and \(y_0\) are:
\[ t = 2 \, \text{s}, \quad y_0 = 18 \, \text{m} \]