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 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
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For position functions of the form \(x(t) = t e^{-t/\tau}\), use the product rule to differentiate and find velocity.
Updated On: Jun 19, 2026
- \(-\frac{1}{e}~\text{m/s}\)
- \(-\frac{1}{e^2}~\text{m/s}\)
- \(-\frac{2}{e}~\text{m/s}\)
- \(-\frac{2}{e^2}~\text{m/s}\)
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The Correct Option is B
Solution and Explanation
Step 1: Determine velocity formula.
Velocity is the derivative of position with respect to time: \[ v(t) = \frac{dx}{dt} = \frac{d}{dt} (\alpha t e^{-t/\tau}) \]
Step 2: Differentiate.
Using product rule: \[ v(t) = \alpha e^{-t/\tau} + \alpha t \cdot \left(-\frac{1}{\tau} e^{-t/\tau}\right) = \alpha e^{-t/\tau} \left(1 - \frac{t}{\tau}\right) \]
Step 3: Substitute values at t = 2 s.
\[ v(2) = 1 \cdot e^{-2/1} \left(1 - \frac{2}{1}\right) = e^{-2}(-1) = -\frac{1}{e^2}~\text{m/s} \]
Step 4: Conclusion.
Hence, the velocity at t = 2 s is \(-\frac{1}{e^2}~\text{m/s}\).
Velocity is the derivative of position with respect to time: \[ v(t) = \frac{dx}{dt} = \frac{d}{dt} (\alpha t e^{-t/\tau}) \]
Step 2: Differentiate.
Using product rule: \[ v(t) = \alpha e^{-t/\tau} + \alpha t \cdot \left(-\frac{1}{\tau} e^{-t/\tau}\right) = \alpha e^{-t/\tau} \left(1 - \frac{t}{\tau}\right) \]
Step 3: Substitute values at t = 2 s.
\[ v(2) = 1 \cdot e^{-2/1} \left(1 - \frac{2}{1}\right) = e^{-2}(-1) = -\frac{1}{e^2}~\text{m/s} \]
Step 4: Conclusion.
Hence, the velocity at t = 2 s is \(-\frac{1}{e^2}~\text{m/s}\).
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