The position x of a particle varies with time t as `x = 6 + 12t - 2 t^2` where x is in metre and t is in seconds. The distance travelled by the particle in first five seconds is

Options

A. 16 m
B. 26 m
C. 10 m
D. 36 m

Explanation

To find the distance traveled by the particle in the first five seconds, we need to follow these steps:

### Step 1: Determine the position function
The position of the particle is given by the equation:
\[ x(t) = 6 + 12t - 2t^2 \]

### Step 2: Find the velocity function
To find the points where the particle changes direction, we first need to find the velocity by differentiating the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = 12 - 4t \]

### Step 3: Find when the velocity is zero
Set the velocity function to zero to find the time at which the particle stops or changes direction:
\[ 12 - 4t = 0 \]
\[ 4t = 12 \]
\[ t = 3 \, \text{seconds} \]

### Step 4: Calculate the position at key times
Now we will calculate the position of the particle at \( t = 0 \), \( t = 3 \), and \( t = 5 \).

1. **At \( t = 0 \)**:
\[ x(0) = 6 + 12(0) - 2(0)^2 = 6 \, \text{m} \]

2. **At \( t = 3 \)**:
\[ x(3) = 6 + 12(3) - 2(3)^2 = 6 + 36 - 18 = 24 \, \text{m} \]

3. **At \( t = 5 \)**:
\[ x(5) = 6 + 12(5) - 2(5)^2 = 6 + 60 - 50 = 16 \, \text{m} \]

### Step 5: Calculate the distance traveled
Now we can find the distance traveled by the particle in the first five seconds. The particle moves from \( x(0) = 6 \, \text{m} \) to \( x(3) = 24 \, \text{m} \) and then back to \( x(5) = 16 \, \text{m} \).

1. Distance from \( t = 0 \) to \( t = 3 \):
\[ \text{Distance} = x(3) - x(0) = 24 - 6 = 18 \, \text{m} \]

2. Distance from \( t = 3 \) to \( t = 5 \):
\[ \text{Distance} = x(3) - x(5) = 24 - 16 = 8 \, \text{m} \]

### Step 6: Total distance traveled
Now, add the distances from both intervals:
\[ \text{Total Distance} = 18 \, \text{m} + 8 \, \text{m} = 26 \, \text{m} \]

Thus, the distance traveled by the particle in the first five seconds is **26 meters**.

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