If a particle starts moving along a straight ine withinitial velocity u under contact acceleration a, its displacement with time is given by the relation `x=ut+(1)/(2)at^2`
Q. Differentiation of `x` w.r.t. `t` will be

Options

A. `y=(at)/(2)`
B. `u+at`
C. `u+2at`
D. `(ut^2)/(2)+(at^3)/(6)`

Explanation

To find the differentiation of the displacement \( x \) with respect to time \( t \), we start with the given equation:

\[ x = ut + \frac{1}{2} a t^2 \]

### Step 1: Differentiate \( x \) with respect to \( t \)

We will differentiate both terms on the right-hand side of the equation separately.

\[ \frac{dx}{dt} = \frac{d}{dt}(ut) + \frac{d}{dt}\left(\frac{1}{2} a t^2\right) \]

### Step 2: Differentiate the first term \( ut \)

Since \( u \) is a constant (initial velocity), we can apply the constant multiple rule:

\[ \frac{d}{dt}(ut) = u \cdot \frac{d}{dt}(t) = u \cdot 1 = u \]

### Step 3: Differentiate the second term \( \frac{1}{2} a t^2 \)

Here, \( \frac{1}{2} a \) is also a constant (as \( a \) is constant acceleration), so we apply the power rule:

\[ \frac{d}{dt}\left(\frac{1}{2} a t^2\right) = \frac{1}{2} a \cdot \frac{d}{dt}(t^2) = \frac{1}{2} a \cdot 2t = a t \]

### Step 4: Combine the results

Now, we can combine the results from the differentiation of both terms:

\[ \frac{dx}{dt} = u + a t \]

### Final Result

Thus, the differentiation of \( x \) with respect to \( t \) is:

\[ \frac{dx}{dt} = u + a t \]