The distance moved by a particle travelling in a straight line in `t`  seconds is given by `s=45t+11t^2-t^3` . The time taken by the particle to come to rest is

Options

A. (a) `9 sec`
B. (b) `5/3 sec`
C. (c) `3/5
sec`
D. (d) `2 sec`

Explanation

To find the time taken by the particle to come to rest, we need to determine when its velocity becomes zero. The velocity of the particle is the derivative of the distance function with respect to time.

### Step 1: Write the distance function
The distance moved by the particle is given by:
\[ s(t) = 45t + 11t^2 - t^3 \]

### Step 2: Find the velocity function
To find the velocity, we take the derivative of the distance function \( s(t) \):
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(45t + 11t^2 - t^3) \]
Using the power rule for differentiation:
\[ v(t) = 45 + 22t - 3t^2 \]

### Step 3: Set the velocity to zero
To find the time when the particle comes to rest, we set the velocity function to zero:
\[ 45 + 22t - 3t^2 = 0 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ -3t^2 + 22t + 45 = 0 \]
Multiplying through by -1 to make it easier to work with:
\[ 3t^2 - 22t - 45 = 0 \]

### Step 5: Use the quadratic formula
To solve for \( t \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 3 \), \( b = -22 \), and \( c = -45 \).

### Step 6: Calculate the discriminant
First, calculate the discriminant:
\[ b^2 - 4ac = (-22)^2 - 4 \cdot 3 \cdot (-45) = 484 + 540 = 1024 \]

### Step 7: Substitute into the quadratic formula
Now substitute back into the quadratic formula:
\[ t = \frac{22 \pm \sqrt{1024}}{2 \cdot 3} = \frac{22 \pm 32}{6} \]

### Step 8: Calculate the possible values for \( t \)
Calculating the two possible values:
1. \( t = \frac{22 + 32}{6} = \frac{54}{6} = 9 \)
2. \( t = \frac{22 - 32}{6} = \frac{-10}{6} = -\frac{5}{3} \) (not physically meaningful since time cannot be negative)

### Conclusion
The time taken by the particle to come to rest is:
\[ t = 9 \text{ seconds} \]