A particle starts from origin with a velocity (in m/s) given by the equation `(dx)/(dt) = x+ 1`. The time (in second) taken by the particle to traverse a distance of 24 m is

Options

A. ln 24
B. ln 5
C. 2 ln 5
D. 2 ln 4

Answer

To solve the problem step by step, we need to find the time taken by the particle to traverse a distance of 24 m given the velocity equation \( \frac{dx}{dt} = x + 1 \).

### Step 1: Set up the equation
We start with the equation of velocity:
\[ \frac{dx}{dt} = x + 1 \]

### Step 2: Separate variables
We can rearrange the equation to separate the variables \( x \) and \( t \):
\[ \frac{dx}{x + 1} = dt \]

### Step 3: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dx}{x + 1} = \int dt \]
The left side integrates to \( \ln|x + 1| \) and the right side integrates to \( t + C \):
\[ \ln|x + 1| = t + C \]

### Step 4: Solve for the constant \( C \)
We know that the particle starts from the origin, which means at \( t = 0 \), \( x = 0 \):
\[ \ln|0 + 1| = 0 + C \implies \ln(1) = C \implies C = 0 \]
Thus, the equation simplifies to:
\[ \ln|x + 1| = t \]

### Step 5: Solve for \( x \)
Now, we can express \( x \) in terms of \( t \):
\[ x + 1 = e^t \implies x = e^t - 1 \]

### Step 6: Find the time when \( x = 24 \)
We need to find the time \( t \) when the particle has traversed a distance of 24 m:
\[ 24 = e^t - 1 \implies e^t = 25 \implies t = \ln(25) \]

### Step 7: Simplify the answer
We can express \( \ln(25) \) as:
\[ \ln(25) = \ln(5^2) = 2\ln(5) \]

### Final Answer
Thus, the time taken by the particle to traverse a distance of 24 m is:
\[ t = 2\ln(5) \text{ seconds} \]

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