Question

A point moves in a straight line under the retardation $-$kv, where k is a constant. If the initial velocity is u, the distance covered in 't' seconds is?

• A. $\frac{u}{k}(1-e^{-kt})$
• B. $\frac{u}{k}(1+e^{-kt})$
• C. $\frac{u}{k}(1-e^{kt})$
• D. $\frac{u}{k}(1+e^{kt})$

Answer 1

Answer: (A)

$\frac{u}{k}(1-e^{-kt})$

$a_d=-KV\frac{dV}{dt}=-KV{\int }_{u_0}^{v_0}\frac{dV}{dt}=-K\int dt\stackrel{v_0}{\underset{u_0}{\left[\ln V\right]=-Kt}}\ln \frac{v_0}{u_0}=-Kt\frac{v_0}{u_0}=e^{-Kt}{v}_0=u_0{e}^{-Kt}v=u{e}^{-Kt}a=V\frac{dV}{dx}kV=V\frac{dV}{dx}k{\int }_0^{x}dx={\int }_u^{v_0}dVkx=v_0-u=u(1-e^{-Kt})x=\frac{u}{K}(1-e^{-Kt})$