A particle is moving along the X-axis with ve | vedclass

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

Question

(D) The velocity of the particle is given by $v = e^{-\beta x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = e^{-\beta x}$.Rearranging the terms

Vedclass · Accepted Answer

$\frac{1}{\beta} \log [1+\beta t]$

Vedclass · Answer

(D) The velocity of the particle is given by $v = e^{-\beta x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = e^{-\beta x}$.Rearranging the terms to separate the variables,we get $e^{\beta x} dx = dt$.Integrating both sides with the initial condition that at $t = 0$,$x = 0$:$\int_0^x e^{\beta x} dx = \int_0^t dt$$\left[ \frac{e^{\beta x}}{\beta} \right]_0^x = [t]_0^t$$\frac{1}{\beta} (e^{\beta x} - e^0) = t - 0$$\frac{1}{\beta} (e^{\beta x} - 1) = t$$e^{\beta x} - 1 = \beta t$$e^{\beta x} = 1 + \beta t$Taking the natural logarithm on both sides:$\beta x = \log(1 + \beta t)$$x = \frac{1}{\beta} \log(1 + \beta t)$

$A$ particle is moving along the $X$-axis with velocity $v = e^{-\beta x}$. At time $t = 0$,the particle is located at $x = 0$. The displacement of the particle as a function of time is

Similar Questions

The motion of a particle is described by the equation $v = at$. The distance travelled by the particle in the first $4 \ s$ is: (in $a$)

If a body having initial velocity zero is moving with uniform acceleration $8\,m/s^2$,the distance travelled by it in the fifth second will be.........$m$.

If average and instantaneous velocities are same,then what is the type of motion?

Derive the equations of uniformly accelerated motion by the graphical method.

$A$ body is moving from rest under constant acceleration. Let $S_1$ be the displacement in the first $(p - 1) \ s$ and $S_2$ be the displacement in the first $p \ s$. The displacement in the $(p^2 - p + 1)^{th} \ s$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module