A particle moves along a straight line AB. At a time t (in seconds) the distance x (in metres) of the particle from O is given by x = 600 + 12t – `t^3`. How long would the particle travel before coming to rest: -

Options

A. 24m
B. 40m
C. 56m
D. 16m

Explanation

To solve the problem step by step, we will analyze the motion of the particle based on the given equation for displacement.

### Step-by-Step Solution:

1. **Understand the equation of motion**:
The displacement \( x \) of the particle is given by the equation:
\[ x = 600 + 12t - t^3 \]
where \( x \) is in meters and \( t \) is in seconds.

2. **Find the velocity**:
The velocity \( v \) of the particle is the derivative of the displacement \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(600 + 12t - t^3) \]
Calculating the derivative:
\[ v = 0 + 12 - 3t^2 = 12 - 3t^2 \]

3. **Set the velocity to zero to find when the particle comes to rest**:
To find the time when the particle comes to rest, we set the velocity \( v \) equal to zero:
\[ 12 - 3t^2 = 0 \]
Rearranging gives:
\[ 3t^2 = 12 \]
Dividing both sides by 3:
\[ t^2 = 4 \]
Taking the square root:
\[ t = 2 \text{ seconds} \]
(We only consider the positive value since time cannot be negative.)

4. **Calculate the position of the particle at \( t = 2 \) seconds**:
Substitute \( t = 2 \) into the displacement equation:
\[ x(2) = 600 + 12(2) - (2)^3 \]
Simplifying:
\[ x(2) = 600 + 24 - 8 = 616 \text{ meters} \]

5. **Calculate the initial position of the particle at \( t = 0 \)**:
Substitute \( t = 0 \) into the displacement equation:
\[ x(0) = 600 + 12(0) - (0)^3 = 600 \text{ meters} \]

6. **Calculate the distance traveled before coming to rest**:
The distance traveled by the particle before coming to rest is:
\[ \text{Distance} = x(2) - x(0) = 616 - 600 = 16 \text{ meters} \]

### Final Answer:
The particle travels **16 meters** before coming to rest.