Find out the position of particle as a function of time which is moving in a straight line with an acceleration `a=5 x`, where x is position.
(i) At `t=0` the particle is at `x=0` (ii) Mass of the object is `5 kg`

Options

A. Question can be solved by information I only.
B. Question can be solved by information II only
C. Question can be solved by information I and II in combined form only.
D. Question can not be solved by both the informations only.

Answer

To find the position of a particle as a function of time when it is moving in a straight line with an acceleration given by \( a = 5x \), we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and position
The acceleration \( a \) can be expressed in terms of velocity \( v \) and position \( x \) as:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can rewrite this as:
\[ a = v \frac{dv}{dx} \]

### Step 2: Set up the equation
Given that \( a = 5x \), we can substitute this into our equation:
\[ v \frac{dv}{dx} = 5x \]

### Step 3: Separate variables and integrate
We can separate the variables:
\[ v \, dv = 5x \, dx \]
Now, we integrate both sides:
\[ \int v \, dv = \int 5x \, dx \]
This gives us:
\[ \frac{v^2}{2} = \frac{5x^2}{2} + C \]
where \( C \) is the integration constant.

### Step 4: Solve for velocity
Multiplying through by 2, we have:
\[ v^2 = 5x^2 + 2C \]
Taking the square root of both sides, we find:
\[ v = \sqrt{5x^2 + 2C} \]

### Step 5: Relate velocity to time
Recall that velocity \( v \) is also defined as:
\[ v = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = \sqrt{5x^2 + 2C} \]

### Step 6: Separate variables again
We can separate the variables again:
\[ \frac{dx}{\sqrt{5x^2 + 2C}} = dt \]

### Step 7: Integrate to find position as a function of time
Now we need to integrate both sides. This integral can be complex, but we can denote it as:
\[ \int \frac{dx}{\sqrt{5x^2 + 2C}} = \int dt \]
Let \( C_2 \) be the constant of integration for the time side.

### Step 8: Apply initial conditions
At \( t = 0 \), \( x = 0 \). We can use this to find the constant \( C \) and \( C_2 \). However, we have two constants and only one initial condition, which means we cannot uniquely determine the function without additional information.

### Conclusion
Thus, the position of the particle as a function of time cannot be uniquely determined with the given information.

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