Question

For a body moving along x-axis, its velocity is given by $v=\frac{a}{b}(1-e^{-bt})$, where a and b are positive constant. At $t=0$, $x=0$. Position x of the particle with time t is given by?

• A. $\frac{a}{b}(1-e^{-bt})t$
• B. $\frac{a}{b}[t+\frac{e^{-bt}}{b}]$
• C. $\frac{a}{b}[t+\frac{1}{b}(e^{-bt}-1\left)\right]$
• D. $\frac{a}{b}[t-\frac{1}{b}(e^{-bt}-1\left)\right]$

Answer 1

The correct option is B $\frac{a}{b}[t+\frac{e^{-bt}}{b}]$

Given velocity $v=\frac{a}{b}(1-e^{-bt})$

At $t=0displacementx=0$

also displacement$x=velocity\left(v\right)$\times$time\left(t\right)$

Thus integrating the velocity v with respect to time we get the relation for displacement.

$vdt=\frac{a}{b}(1-e^{-bt})dt$

$vdt=\frac{a}{b}(t-\frac{e^{-bt}}{-b})$

$=\frac{a}{b}(t+\frac{e^{-bt}}{b})$