The position of a particle moving along x-axis is given by `x = 10t - 2t^(2)`. Then the time (t) at which it will momentily come to rest is

Options

A. Zero
B. 2.5 s
C. 5 s
D. 10 s

Answer

To solve the problem, we need to find the time at which the particle comes to rest. The position of the particle is given by the equation:

\[ x(t) = 10t - 2t^2 \]

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

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(10t - 2t^2) \]

### Step 2: Differentiate the position function
Now, we differentiate the position function:

\[ v(t) = \frac{d}{dt}(10t) - \frac{d}{dt}(2t^2) \]

Calculating the derivatives:

- The derivative of \( 10t \) is \( 10 \).
- The derivative of \( 2t^2 \) is \( 4t \).

Putting it together, we have:

\[ v(t) = 10 - 4t \]

### Step 3: Set the velocity to zero
To find the time at which the particle comes to rest, we set the velocity equal to zero:

\[ 10 - 4t = 0 \]

### Step 4: Solve for \( t \)
Now, we solve for \( t \):

\[ 4t = 10 \]
\[ t = \frac{10}{4} = 2.5 \text{ seconds} \]

### Conclusion
The time at which the particle momentarily comes to rest is:

\[ t = 2.5 \text{ seconds} \]