A particle moves along a straight line `OX`. At a time `t` (in seconds) the distance `x` (in metre) of the particle is given by `x = 40 +12 t - t^3`. How long would the particle travel before coming to rest ?

Options

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

Explanation

To solve the problem, we need to find out how long the particle travels before coming to rest. We are given the position function of the particle as:

\[ x(t) = 40 + 12t - t^3 \]

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

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(40 + 12t - t^3) \]

Calculating the derivative:

\[ v(t) = 0 + 12 - 3t^2 = 12 - 3t^2 \]

### Step 2: Set the velocity to zero to find when the particle comes to rest
The particle comes to rest when its velocity is zero:

\[ 0 = 12 - 3t^2 \]

Rearranging the equation gives:

\[ 3t^2 = 12 \]

Dividing both sides by 3:

\[ t^2 = 4 \]

Taking the square root of both sides:

\[ t = 2 \quad (\text{since time cannot be negative}) \]

### Step 3: Find the distance traveled before coming to rest
Now, we need to find the distance traveled by the particle from \( t = 0 \) to \( t = 2 \).

1. **Calculate the position at \( t = 0 \)**:

\[ x(0) = 40 + 12(0) - (0)^3 = 40 \text{ meters} \]

2. **Calculate the position at \( t = 2 \)**:

\[ x(2) = 40 + 12(2) - (2)^3 \]

Calculating this gives:

\[ x(2) = 40 + 24 - 8 = 56 \text{ meters} \]

### Step 4: Calculate the distance traveled
The distance traveled by the particle before coming to rest is:

\[ \text{Distance} = x(2) - x(0) = 56 - 40 = 16 \text{ meters} \]

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

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