An object is given an initial velocity of 11m/s towards the north and a constant acceleration of `2m//s^(2)` towards the south. What will be the distance covered by the object in the sixth second of its motion

Options

A. 0.25m
B. 0.5m
C. 0
D. 2m

Explanation

To solve the problem of finding the distance covered by an object in the sixth second of its motion, we can follow these steps:

### Step 1: Identify the Initial Conditions
- The object has an initial velocity \( u = 11 \, \text{m/s} \) towards the north.
- The object experiences a constant acceleration \( a = -2 \, \text{m/s}^2 \) (negative because it is directed towards the south).

### Step 2: Determine When the Object Stops
Using the first equation of motion:
\[ v = u + at \]
Where:
- \( v \) is the final velocity (which will be 0 when the object stops),
- \( u = 11 \, \text{m/s} \),
- \( a = -2 \, \text{m/s}^2 \).

Setting \( v = 0 \):
\[ 0 = 11 - 2t \]
Solving for \( t \):
\[ 2t = 11 \implies t = \frac{11}{2} = 5.5 \, \text{s} \]
This means the object will stop after 5.5 seconds.

### Step 3: Analyze Motion in the Sixth Second
We need to find the distance covered during the sixth second, which is the distance traveled from \( t = 5 \, \text{s} \) to \( t = 6 \, \text{s} \).

### Step 4: Calculate Distance Traveled in the Sixth Second
To find the distance traveled in the sixth second, we can find the distance traveled in the first 6 seconds and the distance traveled in the first 5 seconds, then subtract the two.

1. **Distance in the first 6 seconds**:
Using the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
For \( t = 6 \, \text{s} \):
\[ S_6 = 11(6) + \frac{1}{2}(-2)(6^2) \]
Calculating:
\[ S_6 = 66 - \frac{1}{2}(2)(36) = 66 - 36 = 30 \, \text{m} \]

2. **Distance in the first 5 seconds**:
For \( t = 5 \, \text{s} \):
\[ S_5 = 11(5) + \frac{1}{2}(-2)(5^2) \]
Calculating:
\[ S_5 = 55 - \frac{1}{2}(2)(25) = 55 - 25 = 30 \, \text{m} \]

### Step 5: Calculate Distance in the Sixth Second
The distance covered in the sixth second is:
\[ \text{Distance in sixth second} = S_6 - S_5 = 30 \, \text{m} - 30 \, \text{m} = 0 \, \text{m} \]

### Step 6: Finding the Distance Traveled in the Sixth Second
Since the object changes direction during the sixth second, we need to calculate the distance traveled during that time.

1. **Distance traveled from \( t = 5.5 \) to \( t = 6 \)**:
Using the formula for distance:
\[ S = ut + \frac{1}{2} a t^2 \]
Where \( u = 0 \) at \( t = 5.5 \) and \( t = 0.5 \) seconds later:
\[ S_{5.5 \text{ to } 6} = 0(0.5) + \frac{1}{2}(2)(0.5^2) = 0 + \frac{1}{2}(2)(0.25) = 0.25 \, \text{m} \]

2. **Total Distance in the Sixth Second**:
The object travels \( 0.25 \, \text{m} \) upwards in the first half of the sixth second and then \( 0.25 \, \text{m} \) downwards in the second half. Thus, the total distance covered in the sixth second is:
\[ \text{Total Distance} = 0.25 + 0.25 = 0.5 \, \text{m} \]

### Final Answer
The distance covered by the object in the sixth second of its motion is \( 0.5 \, \text{m} \).