A particle moves in a straight line with a velocity given by `(dx)/(dt)=x+1` (x is the distance described).
The time taken by the particle to traverse a distance of 99 metres is

Options

A. `log_(10)e `
B. `2log_(e )10`
C. `2log_(10)e`
D. `(1)/(2)log_(10)e`

Answer

To solve the problem of finding the time taken by a particle to traverse a distance of 99 meters given the velocity equation \(\frac{dx}{dt} = x + 1\), we will follow these steps:

### Step 1: Separate the Variables
We start with the equation:
\[ \frac{dx}{dt} = x + 1 \]
We can rearrange this to separate the variables:
\[ \frac{dx}{x + 1} = dt \]

### Step 2: Integrate Both Sides
Next, we integrate both sides. The left side will be integrated with respect to \(x\) and the right side with respect to \(t\):
\[ \int \frac{dx}{x + 1} = \int dt \]
The integral of the left side is:
\[ \ln|x + 1| + C_1 \]
And the integral of the right side is:
\[ t + C_2 \]
Combining these gives:
\[ \ln|x + 1| = t + C \]
where \(C = C_2 - C_1\) is a constant.

### Step 3: Solve for the Constant \(C\)
To find the constant \(C\), we can use the initial condition. When \(t = 0\), we assume the particle starts from \(x = 0\):
\[ \ln|0 + 1| = 0 + C \implies \ln(1) = C \implies C = 0 \]
Thus, our equation simplifies to:
\[ \ln|x + 1| = t \]

### Step 4: Solve for Time \(t\) when \(x = 99\)
Now we need to find the time \(t\) when the distance \(x\) is 99 meters:
\[ \ln|99 + 1| = t \implies \ln(100) = t \]

### Step 5: Simplify the Result
Using properties of logarithms, we can express \(\ln(100)\):
\[ \ln(100) = \ln(10^2) = 2\ln(10) \]
Thus, the time taken by the particle to traverse 99 meters is:
\[ t = 2\ln(10) \]

### Final Answer
The time taken by the particle to traverse a distance of 99 meters is:
\[ t = 2 \ln(10) \]

---