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 find the time at which the particle comes to rest, we need to determine when its velocity is zero. The position of the particle is given by the equation:

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

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

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

### Step 2: Differentiate the position function
Using the power rule for differentiation:

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

Thus, the velocity function is:

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

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

\[ 10 - 4t = 0 \]

### Step 4: Solve for \( t \)
Rearranging the equation gives:

\[ 4t = 10 \]

Dividing both sides by 4:

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

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

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

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