**Given Information:**
The distance covered by a body moving in a straight line in time t is given by the relation (2x)^2 - 3x = t.

**Determining Velocity:**
To find the velocity of the body at a certain instant of time, we need to differentiate the given equation with respect to time (t).

Differentiating the equation (2x)^2 - 3x = t with respect to t, we get:

4(2x)(dx/dt) - 3(dx/dt) = 1

Simplifying the equation, we have:

8x(dx/dt) - 3(dx/dt) = 1

Factoring out (dx/dt), we get:

(8x - 3)(dx/dt) = 1

Dividing both sides by (8x - 3), we get:

dx/dt = 1 / (8x - 3)

Therefore, the velocity of the body at a certain instant of time is given by:

v = dx/dt = 1 / (8x - 3)

**Determining Acceleration:**
To find the acceleration of the body, we need to differentiate the velocity equation with respect to time (t).

Differentiating the velocity equation v = 1 / (8x - 3) with respect to t, we get:

dv/dt = d(1 / (8x - 3)) / dt

To differentiate 1 / (8x - 3), we can use the chain rule. Let u = 8x - 3.

dv/dt = d(1 / u) / dt

Using the chain rule, we have:

dv/dt = (-1 / u^2) * du/dt

Substituting u = 8x - 3, we get:

dv/dt = (-1 / (8x - 3)^2) * d(8x - 3) / dt

Simplifying the equation, we have:

dv/dt = (-1 / (8x - 3)^2) * (8(dx/dt))

Since dx/dt represents the velocity, we can substitute it with v:

dv/dt = (-8v / (8x - 3)^2)

Therefore, the acceleration of the body at a certain instant of time is given by:

a = dv/dt = (-8v / (8x - 3)^2)

**Conclusion:**
The acceleration of the body at a certain instant of time is given by the equation a = (-8v / (8x - 3)^2), where v is the velocity of the body.

-4v*3