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

Options

A. `(2s,3s)`
B. `(3s,4s)`
C. `(4s,8s)`
D. `(1s,2s)`

Explanation

To solve the problem, we can use the equations of motion under uniform acceleration. The relevant equation for this situation is:

\[ x = x_0 + v_0 t + \frac{1}{2} a t^2 \]

Where:
- \( x \) is the final position (12 m)
- \( x_0 \) is the initial position (0 m)
- \( v_0 \) is the initial velocity (10 m/s)
- \( a \) is the acceleration (-4 m/s²)
- \( t \) is the time taken to reach the position x

### Step 1: Substitute the known values into the equation
We know:
- \( x = 12 \, \text{m} \)
- \( x_0 = 0 \, \text{m} \)
- \( v_0 = 10 \, \text{m/s} \)
- \( a = -4 \, \text{m/s}^2 \)

Substituting these values into the equation gives us:

\[ 12 = 0 + 10t + \frac{1}{2}(-4)t^2 \]

### Step 2: Simplify the equation
This simplifies to:

\[ 12 = 10t - 2t^2 \]

Rearranging gives us a standard quadratic equation:

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

### Step 3: Divide the entire equation by 2
To make calculations easier, we can divide the entire equation by 2:

\[ t^2 - 5t + 6 = 0 \]

### Step 4: Factor the quadratic equation
Next, we can factor the quadratic equation:

\[ (t - 2)(t - 3) = 0 \]

### Step 5: Solve for t
Setting each factor to zero gives us the possible solutions for t:

1. \( t - 2 = 0 \) → \( t = 2 \, \text{s} \)
2. \( t - 3 = 0 \) → \( t = 3 \, \text{s} \)

### Step 6: Determine the valid time
Since the body is moving with a negative acceleration (deceleration), we need to consider the time it takes to reach the position of 12 m. Both times (2 s and 3 s) are valid, but we need to check which one is appropriate based on the context of the motion.

Since the body starts at 10 m/s and is decelerating, it will take longer to reach 12 m as it slows down. Therefore, the valid time is:

**Final Answer: \( t = 3 \, \text{s} \)**

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