Question

The velocity of a particle is only a function of its position \(v(x) = e^{-x\). If at \(t=0\), \(x=0\), find displacement as a function of time.}

• A. \(e^{t}\)
• B. \(e^{(1-t)}\)
• C. \(\ln(t)\)
• D. \(\ln(1+t)\)

Hint:

When \(v = f(x)\), always convert into separable differential equation \(dx/f(x) = dt\).

Answer 1

The Correct Option is D

Step 1: Understanding the nature of velocity function.
Here velocity is given as a function of displacement \(x\), not time. This means we cannot directly integrate with respect to time. Instead, we must treat it as a differential equation of the form \( \frac{dx}{dt} = e^{-x} \). This type of problem requires separation of variables.

Step 2: Writing the differential equation properly.
We rewrite the equation as \( \frac{dx}{dt} = e^{-x} \). To solve, we bring all \(x\)-dependent terms to one side and time terms to the other side. This gives a separable structure, which is essential for integration.

Step 3: Separating variables carefully.
We multiply both sides by \(e^{x}\) to isolate variables. This gives \( e^{x} dx = dt \). Now each side contains only one variable, making integration possible. This is a standard technique in velocity-position dependent motion.

Step 4: Performing integration.
We integrate both sides: \( \int e^{x} dx = \int dt \). The left side integrates to \( e^{x} \), and the right side integrates to \( t + C \). Thus we obtain the relation \( e^{x} = t + C \).

Step 5: Applying initial conditions.
At \( t = 0 \), it is given that \( x = 0 \). Substituting into the equation gives \( e^{0} = 0 + C \). Since \( e^{0} = 1 \), we get \( C = 1 \). This step fixes the constant of integration.

Step 6: Final equation formation.
Substituting \( C = 1 \) back, we get \( e^{x} = t + 1 \). Taking natural logarithm on both sides gives \( x = \ln(1 + t) \). This gives displacement as a function of time.

Step 7: Final conclusion.
Thus, the displacement of the particle as a function of time is \( x(t) = \ln(1+t) \).