The displacement of a particle moving in a straight line is described by the relation `s=6+12t-2t^(2)`. Here `s` is in metre and `t` in second. The distance covered by the particle in first `5s` is

Options

A. `20m`
B. `32m`
C. `24m`
D. `26m`

Answer

To solve the problem, we need to find the distance covered by the particle in the first 5 seconds given its displacement function \( s = 6 + 12t - 2t^2 \).

### Step 1: Find the velocity function
The velocity \( v \) of the particle is the derivative of the displacement \( s \) with respect to time \( t \).

\[ v = \frac{ds}{dt} = \frac{d}{dt}(6 + 12t - 2t^2) = 12 - 4t \]

### Step 2: Determine when the velocity is zero
To find when the particle changes direction, we set the velocity function to zero:

\[ 12 - 4t = 0 \]
\[ 4t = 12 \implies t = 3 \text{ seconds} \]

### Step 3: Calculate displacement at \( t = 0 \)
Now we calculate the displacement at \( t = 0 \):

\[ s(0) = 6 + 12(0) - 2(0)^2 = 6 \text{ meters} \]

### Step 4: Calculate displacement at \( t = 3 \)
Next, we calculate the displacement at \( t = 3 \):

\[ s(3) = 6 + 12(3) - 2(3)^2 = 6 + 36 - 18 = 24 \text{ meters} \]

### Step 5: Calculate displacement at \( t = 5 \)
Now we calculate the displacement at \( t = 5 \):

\[ s(5) = 6 + 12(5) - 2(5)^2 = 6 + 60 - 50 = 16 \text{ meters} \]

### Step 6: Calculate the distance covered
The distance covered by the particle can be calculated in two segments:
1. From \( t = 0 \) to \( t = 3 \):
- Initial position: \( s(0) = 6 \) meters
- Position at \( t = 3 \): \( s(3) = 24 \) meters
- Distance covered: \( 24 - 6 = 18 \) meters

2. From \( t = 3 \) to \( t = 5 \):
- Position at \( t = 3 \): \( s(3) = 24 \) meters
- Position at \( t = 5 \): \( s(5) = 16 \) meters
- Distance covered: \( 24 - 16 = 8 \) meters (the particle moves back)

### Step 7: Total distance covered
Now, we can sum the distances covered in both segments:

\[ \text{Total distance} = 18 \text{ meters} + 8 \text{ meters} = 26 \text{ meters} \]

Thus, the total distance covered by the particle in the first 5 seconds is **26 meters**.

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