The acceleration a of `alpha` of a particle depends on displacements covered in a time t as `a= S + 5` . It is given that initially `S=0` m and `v=5m//s`. Then

Options

A. `v= S+5`
B. `v= sqrt(S+5)`
C. `t= log_(e)((S+5)/(S))`
D. `t= log_(e)((S+5)/(5))`

Answer

To solve the problem, we need to derive expressions for velocity and time in terms of displacement \( S \) given the relationship between acceleration and displacement.

### Step 1: Understand the relationship between acceleration, velocity, and displacement
We are given the equation for acceleration:
\[ a = S + 5 \]
Acceleration can also be expressed in terms of velocity and displacement:
\[ a = \frac{dv}{dt} = v \frac{dv}{ds} \]
where \( v \) is the velocity and \( s \) is the displacement.

### Step 2: Set up the equation
Substituting the expression for acceleration into the equation gives us:
\[ v \frac{dv}{ds} = S + 5 \]
Rearranging this, we have:
\[ v dv = (S + 5) ds \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side integrates with respect to \( v \) and the right side with respect to \( s \):
\[ \int v \, dv = \int (S + 5) \, ds \]
This results in:
\[ \frac{v^2}{2} = \frac{S^2}{2} + 5S + C \]
where \( C \) is the constant of integration.

### Step 4: Apply initial conditions
We know that initially \( S = 0 \) and \( v = 5 \) m/s. Plugging these values into the equation:
\[ \frac{5^2}{2} = \frac{0^2}{2} + 5(0) + C \]
This simplifies to:
\[ \frac{25}{2} = C \]
So, the equation becomes:
\[ \frac{v^2}{2} = \frac{S^2}{2} + 5S + \frac{25}{2} \]

### Step 5: Simplify the equation
Multiplying through by 2 to eliminate the fractions:
\[ v^2 = S^2 + 10S + 25 \]
This can be factored as:
\[ v^2 = (S + 5)^2 \]
Taking the square root gives us:
\[ v = S + 5 \]

### Step 6: Find the expression for time
Now we need to find the expression for time \( t \) in terms of displacement \( S \). We know:
\[ v = \frac{ds}{dt} = S + 5 \]
Rearranging gives us:
\[ dt = \frac{ds}{S + 5} \]

### Step 7: Integrate to find time
Integrating both sides:
\[ \int dt = \int \frac{ds}{S + 5} \]
This results in:
\[ t = \ln(S + 5) - \ln(5) + C' \]
Using the initial condition where \( S = 0 \) at \( t = 0 \):
\[ 0 = \ln(5) + C' \implies C' = -\ln(5) \]
Thus, the equation for time becomes:
\[ t = \ln\left(\frac{S + 5}{5}\right) \]

### Final expressions
1. The expression for velocity in terms of displacement is:
\[ v = S + 5 \]
2. The expression for time in terms of displacement is:
\[ t = \ln\left(\frac{S + 5}{5}\right) \]