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, we need to analyze the motion of the particle in both the x and y directions separately.

### Step 1: Analyze the motion in the x-direction
The particle starts at the origin (0,0) with an initial velocity of \( u_x = 0 \, \text{m/s} \) (since it only has an initial velocity in the y-direction) and has a constant acceleration \( a_x = 10 \, \text{m/s}^2 \). The displacement in the x-direction is given as \( x = 20 \, \text{m} \).

Using the equation of motion:
\[ x = u_x t + \frac{1}{2} a_x t^2 \]
Substituting the known values:
\[ 20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
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 analyze the motion in the y-direction. The initial velocity in the y-direction is \( u_y = 5 \, \text{m/s} \) and the acceleration in the y-direction is \( a_y = 4 \, \text{m/s}^2 \). We need to find the displacement \( y \) after the time \( t = 2 \, \text{s} \).

Using the equation of motion:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ y = 5 \cdot 2 + \frac{1}{2} \cdot 4 \cdot (2^2) \]
Calculating this gives:
\[ y = 10 + \frac{1}{2} \cdot 4 \cdot 4 \]
\[ y = 10 + 8 \]
\[ y = 18 \, \text{m} \]

### Final Results
The values we have found are:
- Time \( t = 2 \, \text{s} \)
- Displacement in the y-direction \( y = 18 \, \text{m} \)

### Summary
Thus, the final answers are:
- \( t = 2 \, \text{s} \)
- \( y = 18 \, \text{m} \)

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