Question

A particle is moving along straight line. If its location at time t is given by \(x(t) = \alpha t e^{-t/\tau}; \alpha = 1~\text{m/s}, \tau = 1~\text{s}\). The velocity of the particle at t = 2 s is

• A. \(-\frac{1}{e}~\text{m/s}\)
• B. \(-\frac{1}{e^2}~\text{m/s}\)
• C. \(-\frac{2}{e}~\text{m/s}\)
• D. \(-\frac{2}{e^2}~\text{m/s}\)

Hint:

For position functions of the form \(x(t) = t e^{-t/\tau}\), use the product rule to differentiate and find velocity.

Answer 1

The Correct Option is B

Step 1: Velocity as the time derivative.
v(t) = dx/dt = d/dt [α t e^(-t/τ)].

Step 2: Executing differentiation.
Applying the product rule: v(t) = α e^(-t/τ) + α t (-1/τ) e^(-t/τ) = α e^(-t/τ) (1 - t/τ).

Step 3: Plugging in t = 2 s.
v(2) = 1 · e^(-2/1) (1 - 2/1) = e^(-2) (-1) = -1/e² m/s.

Step 4: Conclusion.
Hence, the velocity at t = 2 s is -1/e² m/s.