The acceleration of a particle travelling along a straight line is `a =(k)/(v)`, where `k` is constant `(k gt 0)`. If at `t=0,V=V_(0)` then speed of particle at `t = (V_(0)^(2))/(2)`.

Explanation

To solve the problem, we start with the given acceleration of a particle traveling along a straight line, which is defined as:

\[ a = \frac{k}{v} \]

where \( k \) is a constant and \( v \) is the velocity of the particle. We know that acceleration can also be expressed in terms of velocity as:

\[ a = \frac{dv}{dt} \]

Thus, we can equate the two expressions for acceleration:

\[ \frac{dv}{dt} = \frac{k}{v} \]

Now, we can rearrange this equation to separate the variables \( v \) and \( t \):

\[ v \, dv = k \, dt \]

Next, we will integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):

\[ \int v \, dv = \int k \, dt \]

The integral of \( v \, dv \) is:

\[ \frac{v^2}{2} + C_1 \]

And the integral of \( k \, dt \) is:

\[ kt + C_2 \]

Setting the constants of integration equal to zero for simplicity, we have:

\[ \frac{v^2}{2} = kt \]

Now, we need to apply the limits for the integration. At \( t = 0 \), the velocity \( v = V_0 \). So we can rewrite our equation with limits:

\[ \frac{v^2}{2} - \frac{V_0^2}{2} = kt \]

At \( t = \frac{V_0^2}{2} \):

\[ \frac{v^2}{2} - \frac{V_0^2}{2} = k \left(\frac{V_0^2}{2}\right) \]

Now, simplifying this equation:

\[ \frac{v^2}{2} - \frac{V_0^2}{2} = \frac{k V_0^2}{2} \]

Multiplying through by 2 to eliminate the fraction:

\[ v^2 - V_0^2 = k V_0^2 \]

Rearranging gives us:

\[ v^2 = V_0^2 + k V_0^2 \]

Factoring out \( V_0^2 \):

\[ v^2 = V_0^2 (1 + k) \]

Taking the square root of both sides, we find:

\[ v = V_0 \sqrt{1 + k} \]

Thus, the speed of the particle at time \( t = \frac{V_0^2}{2} \) is:

\[ v = V_0 \sqrt{1 + k} \]