Question

The deceleration experienced by a block under a force is given by $a=-bv$; where a$=$ acceleration, v$=$ velocity'; b$=$constant. If ${}^{\prime }{v}_0^{\prime }$ is magnitude of velocity at $t=0$, find velocity after time 't'.

• A. $v_0$
• B. $\frac{v_0}{2}$
• C. $v_0{e}^{-bt}$
• D. $\frac{v_0}{e^{-bt}}$

Answer 1

The correct option is C $v_0{e}^{-bt}$

Given a=-bt; as we know change in velocity with respect to time=acceleration

$\frac{dv}{dt}=-bv\frac{dv}{v}-bdt$

Now upon integrating we get,

${\int }_{v_0}^v\frac{dv}{v}={\int }_0^t-bdt=>{logv}_{v_0}^v=-b{t}_0^{t}logv-log{v}_0=-btlog\frac{v}{v_0}=-btv=v_0{e}^{-bt}$