Question

The displacement x of a particle varies with time t as x = ae${}^{-\alpha t}$ + be${}^{\beta t}$, Where a, b, $\alpha$ and $\beta$ are positive constants. The velocity of the particle will:

• A. Be independent of $\alpha$ and $\beta$
• B. Go on increasing with time
• C. Drop to zero when $\alpha$ = $\beta$
• D. Go on decreasing with time

Answer 1

The correct option is B Go on increasing with time

$x=a{e}^{-\alpha t}+b{e}^{\beta t}$

Where a,b, $\alpha$, b are Positive constant

$V=\frac{dx}{dt}=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}$

$\therefore \frac{dv}{dt}=a{\alpha }^2{e}^{-\alpha t}+b{\beta }^2{e}^{\beta t}isalways>0$

V is increasing the function of t.

Hence option (B) is correct.