A particle is moving in a straight line and its velocity varies with its displacement as `v=sqrt(4+4s)` m/s. Assume s=0 at t=0. find the displacement of the particle in metres at t=1s.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the relationship between velocity and displacement
The given relationship is:
\[ v = \sqrt{4 + 4s} \]

### Step 2: Square both sides to eliminate the square root
Squaring both sides gives:
\[ v^2 = 4 + 4s \]

### Step 3: Rearrange the equation to express displacement in terms of velocity
Rearranging the equation, we get:
\[ 4s = v^2 - 4 \]
\[ s = \frac{v^2 - 4}{4} \]

### Step 4: Use the relationship between velocity, acceleration, and time
Since velocity is a function of displacement, we can find acceleration. We know that:
\[ v^2 = u^2 + 2as \]
Here, we can compare this with our earlier equation.

### Step 5: Identify initial conditions
At \( s = 0 \) (initial displacement), we have:
\[ v = \sqrt{4 + 4(0)} = \sqrt{4} = 2 \, \text{m/s} \]
Thus, the initial velocity \( u = 2 \, \text{m/s} \).

### Step 6: Find acceleration
From the equation \( v^2 = u^2 + 2as \):
- Initial velocity \( u = 2 \, \text{m/s} \) gives \( u^2 = 4 \).
- From our rearranged equation, we can see that \( 4 = 4 + 4s \) implies \( a = 2 \, \text{m/s}^2 \).

### Step 7: Use the second equation of motion to find displacement at \( t = 1 \, \text{s} \)
Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 2 \, \text{m/s} \), \( a = 2 \, \text{m/s}^2 \), and \( t = 1 \, \text{s} \):
\[ s = 2(1) + \frac{1}{2}(2)(1^2) \]
\[ s = 2 + \frac{1}{2}(2)(1) \]
\[ s = 2 + 1 = 3 \, \text{m} \]

### Final Answer
The displacement of the particle at \( t = 1 \, \text{s} \) is \( 3 \, \text{m} \).

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