If a particle takes `t` second less and acquire a velocity of `vms^(-1)` more in falling through the same disance on two planets where the accelerations due to gravity are `2g` and `8g` respectively, then `v=xgt`. Find value of `x`

Answer

To solve the problem step by step, we will analyze the motion of the particle falling on two different planets with different gravitational accelerations.

### Step 1: Define the Problem
We have two planets with gravitational accelerations of \(2g\) and \(8g\). The particle takes \(t\) seconds less on the second planet and acquires a velocity of \(v\) m/s more than on the first planet. We need to find the value of \(x\) in the equation \(v = xgt\).

### Step 2: Calculate Distance Fallen on Planet 1
For the first planet (with \(a = 2g\)):
- Let the time taken to fall be \(t\).
- Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, the initial velocity \(u = 0\), so:
\[ s = 0 + \frac{1}{2} (2g) t^2 = g t^2 \]
Let's label this as **Equation (1)**.

### Step 3: Calculate Distance Fallen on Planet 2
For the second planet (with \(a = 8g\)):
- The time taken to fall is \(t - t\) (which is \(t - t = 0\), but we need to consider the time taken as \(t - t\)).
- Again using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, the initial velocity \(u = 0\), so:
\[ s = 0 + \frac{1}{2} (8g) (t - t)^2 = 4g (t - t)^2 \]
Let's label this as **Equation (2)**.

### Step 4: Set the Distances Equal
Since the distances fallen on both planets are the same:
\[ g t^2 = 4g (t - t)^2 \]
We can cancel \(g\) from both sides (assuming \(g \neq 0\)):
\[ t^2 = 4 (t - t)^2 \]

### Step 5: Solve for \(t\)
Taking the square root of both sides:
\[ t = 2(t - t) \]
This simplifies to:
\[ t = 2t - 2t \implies t = 2t \]
This indicates that the time taken on the first planet is twice that of the second planet.

### Step 6: Calculate Velocities
Now, we will calculate the velocities acquired on both planets:
- For Planet 1:
\[ v_1 = a_1 t = 2g t \]
- For Planet 2:
\[ v_2 = a_2 (t - t) = 8g (t - t) = 8g t - 8g t \]

### Step 7: Relate the Velocities
According to the problem, the velocity acquired on the second planet is \(v\) m/s more than that on the first planet:
\[ v_2 = v_1 + v \]
Substituting the expressions for \(v_1\) and \(v_2\):
\[ 8g(t - t) = 2g t + v \]
This simplifies to:
\[ v = 8gt - 2gt = 6gt \]

### Step 8: Final Expression
From the equation \(v = xgt\), we can see that:
\[ x = 6 \]

### Conclusion
Thus, the value of \(x\) is:
\[ \boxed{6} \]