A body is moving along the +ve X-axis with uniform acceleration of `-4ms^(2)` Its velocity at x = 0 is `10 ms^(-1)`. The time taken by the body to reach a point at x = 12m is 

Options

A. (25, 3s)
B. (38, 4s)
C. (48,8s)
D. (1s, 28)

Answer

To solve the problem step by step, we will use the kinematic equation for uniformly accelerated motion:

### Step 1: Identify the given values
- Initial velocity (u) = 10 m/s (at x = 0)
- Acceleration (a) = -4 m/s² (uniform acceleration)
- Displacement (s) = 12 m (the point we want to reach)

### Step 2: Write the kinematic equation
The kinematic equation we will use is:
\[ s = ut + \frac{1}{2} a t^2 \]

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ 12 = 10t + \frac{1}{2}(-4)t^2 \]

### Step 4: Simplify the equation
This simplifies to:
\[ 12 = 10t - 2t^2 \]

Rearranging gives:
\[ 2t^2 - 10t + 12 = 0 \]

### Step 5: Divide the entire equation by 2
To make calculations easier, divide the entire equation by 2:
\[ t^2 - 5t + 6 = 0 \]

### Step 6: Factor the quadratic equation
Now, we need to factor the quadratic equation:
\[ (t - 2)(t - 3) = 0 \]

### Step 7: Solve for t
Setting each factor to zero gives us:
1. \( t - 2 = 0 \) → \( t = 2 \) seconds
2. \( t - 3 = 0 \) → \( t = 3 \) seconds

### Step 8: Conclusion
The time taken by the body to reach the point at x = 12 m is either 2 seconds or 3 seconds.