The dispacement of a body is given by `4s=M +2Nt^(4)`, where `M` and `N` are constants.
The velocity of the body at the end of `1 s` from the start is .

Options

A. `2N`
B. `(M+2N)/(4)`
C. `2(M_N)`
D. `(2M+N)/(4)`

Explanation

To solve the problem, we need to find the velocity of the body at the end of 1 second, given the displacement equation. Let's break it down step by step.

### Step 1: Write down the displacement equation
The displacement \( s \) is given by:
\[ 4s = M + 2N t^4 \]
where \( M \) and \( N \) are constants.

### Step 2: Express displacement \( s \)
To find \( s \), we can rearrange the equation:
\[ s = \frac{M + 2N t^4}{4} \]

### Step 3: Differentiate displacement to find velocity
Velocity \( v \) is the derivative of displacement \( s \) with respect to time \( t \):
\[ v = \frac{ds}{dt} \]
Now, we differentiate \( s \):
\[ v = \frac{d}{dt} \left( \frac{M + 2N t^4}{4} \right) \]

### Step 4: Apply the differentiation
Since \( M \) is a constant, its derivative is 0. The derivative of \( 2N t^4 \) is:
\[ \frac{d}{dt}(2N t^4) = 2N \cdot 4t^3 = 8N t^3 \]
Thus, we have:
\[ v = \frac{1}{4} \cdot 8N t^3 = 2N t^3 \]

### Step 5: Calculate the velocity at \( t = 1 \) second
Now, we can find the velocity at \( t = 1 \):
\[ v(1) = 2N (1)^3 = 2N \]

### Final Answer
The velocity of the body at the end of 1 second is:
\[ \boxed{2N} \]

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