Question

The displacement \( x \) of a particle varies with time \( t \), \( x = ae^{-pt} + be^{qt} \), where \( a, b, p, q \) are positive constants. The velocity of the particle will

• A. go on increasing forever
• B. be independent of \( p \) and \( q \)
• C. drop to zero when \( p = q \)
• D. go on decreasing with time

Hint:

Compare exponential terms by analyzing their behavior at large \( t \). Growing exponential terms dominate decaying ones.

Answer 1

The Correct Option is A

To find velocity, we differentiate displacement with respect to time: \[ x(t) = ae^{-pt} + be^{qt} \] \[ v(t) = \frac{dx}{dt} = -ap e^{-pt} + bq e^{qt} \] Now analyze the behavior as \( t \to \infty \): - Since \( e^{-pt} \to 0 \), the term \( -ap e^{-pt} \to 0 \) - Since \( e^{qt} \to \infty \), the term \( bq e^{qt} \to \infty \) Thus, the exponential growth dominates, and: \[ v(t) \to \infty \text{ as } t \to \infty \] Hence, the velocity increases continuously forever.