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) Given displacement $x = a e^{-\alpha t} + b e^{\beta t}$.Velocity $v$ is the rate of change of displacement with respect to time:$v = \frac{dx}{dt

Vedclass · Accepted Answer

Go on increasing with time

Vedclass · Answer

(D) Given displacement $x = a e^{-\alpha t} + b e^{\beta t}$.Velocity $v$ is the rate of change of displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t})$$v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$.To determine how velocity changes with time,we find the acceleration $a_{acc}$:$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(-a\alpha e^{-\alpha t} + b\beta e^{\beta t})$$a_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$.Since $a, b, \alpha, \beta$ are positive constants and the exponential functions $e^{-\alpha t}$ and $e^{\beta t}$ are always positive for any time $t$,the acceleration $a_{acc}$ is always positive.Because the acceleration is positive,the velocity $v$ will go on increasing with time.

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

Similar Questions

In the picture shown,a ball starting from rest rolls down a ramp $AB$,moves along the horizontal bottom $BC$,then goes up a smaller ramp $CD$,and thereafter rolls on the horizontal plane $DE$. Ignore friction and air resistance. Which of the following figures shows the variation in speed with time?

For the displacement-time graph shown in the figure,the ratio of the magnitudes of the (constant) speeds during the first two seconds and the next four seconds is

$A$ velocity-time graph of a body is shown below. The distance covered by the body from $t = 6 \ s$ to $t = 9 \ s$ is: (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module