A body starts from rest with uniform acceleration. Its velocity after 2n second is `v_(0)`. The displacement of the body in last n second is `(3v_(0)n)/(beta)`. Determine the value of `beta`?

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant equations of motion.

### Step 1: Understand the problem
A body starts from rest with uniform acceleration. After \(2n\) seconds, its velocity is \(v_0\). We need to find the displacement of the body during the last \(n\) seconds and relate it to the given expression \(\frac{3v_0 n}{\beta}\).

### Step 2: Find the acceleration
Using the first equation of motion:
\[ v = u + at \]
where:
- \(v\) is the final velocity after \(2n\) seconds,
- \(u\) is the initial velocity (which is \(0\) since the body starts from rest),
- \(a\) is the acceleration,
- \(t\) is the time (\(2n\) seconds).

Substituting the known values:
\[ v_0 = 0 + a(2n) \]
This gives us:
\[ a = \frac{v_0}{2n} \]

### Step 3: Calculate the total displacement in \(2n\) seconds
Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \(s\) is the total displacement after \(2n\) seconds. Substituting the values:
\[ s = 0 \cdot (2n) + \frac{1}{2} \left(\frac{v_0}{2n}\right)(2n)^2 \]
This simplifies to:
\[ s = \frac{1}{2} \left(\frac{v_0}{2n}\right)(4n^2) = \frac{2v_0 n}{2} = n v_0 \]

### Step 4: Calculate the displacement during the last \(n\) seconds
Now, we need to find the displacement during the last \(n\) seconds. The time interval for this part is from \(n\) seconds to \(2n\) seconds.

Using the formula again for the last \(n\) seconds:
\[ s' = u' t' + \frac{1}{2} a (t')^2 \]
where:
- \(u'\) is the initial velocity at \(n\) seconds, which can be found using:
\[ u' = u + at = 0 + \left(\frac{v_0}{2n}\right)(n) = \frac{v_0}{2} \]
- \(t' = n\) seconds.

Now substituting these values:
\[ s' = \left(\frac{v_0}{2}\right)(n) + \frac{1}{2} \left(\frac{v_0}{2n}\right)(n^2) \]
Calculating this gives:
\[ s' = \frac{v_0 n}{2} + \frac{1}{2} \left(\frac{v_0}{2n}\right)(n^2) = \frac{v_0 n}{2} + \frac{v_0 n}{4} = \frac{2v_0 n}{4} + \frac{v_0 n}{4} = \frac{3v_0 n}{4} \]

### Step 5: Relate the displacement to the given expression
According to the problem, the displacement during the last \(n\) seconds is given as:
\[ s' = \frac{3v_0 n}{\beta} \]
Setting the two expressions for \(s'\) equal to each other:
\[ \frac{3v_0 n}{4} = \frac{3v_0 n}{\beta} \]

### Step 6: Solve for \(\beta\)
To find \(\beta\), we can cancel \(3v_0 n\) from both sides (assuming \(v_0 n \neq 0\)):
\[ \frac{1}{4} = \frac{1}{\beta} \implies \beta = 4 \]

### Final Answer
Thus, the value of \(\beta\) is:
\[ \beta = 4 \]