The displacement x of a particle varies with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The displacement is given by $x = a e^{-\alpha t} + b e^{\beta t}$.The velocity $v$ is the time derivative of displacement: $v = \frac{dx}{dt}$.$v

Vedclass · Accepted Answer

go on increasing with time

Vedclass · Answer

(D) The displacement is given by $x = a e^{-\alpha t} + b e^{\beta t}$.The velocity $v$ is the time derivative of displacement: $v = \frac{dx}{dt}$.$v = \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t}) = a(-\alpha) e^{-\alpha t} + b(\beta) e^{\beta t}$.$v = -a \alpha e^{-\alpha t} + b \beta e^{\beta t}$.As time $t$ increases,the term $e^{-\alpha t}$ decreases towards zero,while the term $e^{\beta t}$ increases exponentially. Since $b, \beta > 0$,the term $b \beta e^{\beta t}$ will dominate for large $t$,causing the velocity $v$ to increase with time.

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

Similar Questions

Charge passing through a conductor of cross-section area $A=0.3 ~m^2$ is given by $q=3 t^2+5 t+2$ in coulomb,where $t$ is in second. What is the value of drift velocity at $t=2 ~s$ ? (Given,$n=2 \times 10^{25} / m^3$ )

The binding energy of deuteron is $2.2\,MeV$ and that of $_2^4He$ is $28\,MeV$. If two deuterons are fused to form one $_2^4He$ nucleus, then the energy released is ......... $MeV$.

During the electrolytic reduction of alumina,the reaction at cathode is

The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$,then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$

Persistent nucellus in the seed is known as

Vedclass Test Series

Exam Paper Generator

Online Exam Module