A particle is moving along a straight line OX, At a time t (in seconds) the distance x (in metres) of particle from point O is given by `x=10+6t-3t^(2)`. How long would the particle travel before coming to rest?

Explanation

To solve the problem of how long the particle travels before coming to rest, we can follow these steps:

### Step 1: Understand the equation of motion
The position of the particle as a function of time is given by:
\[ x(t) = 10 + 6t - 3t^2 \]

### Step 2: Find the velocity function
The velocity of the particle is the derivative of the position function with respect to time:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(10 + 6t - 3t^2) \]
Calculating the derivative:
\[ v(t) = 0 + 6 - 6t = 6 - 6t \]

### Step 3: Set the velocity to zero to find when the particle comes to rest
To find when the particle comes to rest, we set the velocity equal to zero:
\[ 6 - 6t = 0 \]
Solving for \( t \):
\[ 6t = 6 \]
\[ t = 1 \text{ second} \]

### Step 4: Calculate the initial and final positions
Now we need to calculate the distance traveled before the particle comes to rest. We will find the position at \( t = 0 \) and \( t = 1 \).

- At \( t = 0 \):
\[ x(0) = 10 + 6(0) - 3(0)^2 = 10 \text{ meters} \]

- At \( t = 1 \):
\[ x(1) = 10 + 6(1) - 3(1)^2 = 10 + 6 - 3 = 13 \text{ meters} \]

### Step 5: Calculate the distance traveled
The distance traveled by the particle before coming to rest is given by:
\[ \text{Distance} = x_{\text{final}} - x_{\text{initial}} = x(1) - x(0) \]
\[ \text{Distance} = 13 - 10 = 3 \text{ meters} \]

### Final Answer
The particle travels **3 meters** before coming to rest.

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