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) $x = \dfrac{u}{k}(1 - {e^{ - kt}})$
(B) $x = \dfrac{u}{k}(1 + {e^{ - kt}})$
(C) $x = \dfrac{u}{k}(1 - {e^{kt}})$
(D) $x = \dfrac{u}{k}(1 + {e^{kt}})$

Answer 1

Hint : As we know, a body travels in a straight line and if it covers an equal distance in equal duration of time, it is an uniform motion. In our case the retardation is -kV. The motion cannot be said to be an uniform motion. The acceleration makes the velocity reach zero after a certain interval of time.
In such cases we use integration methods to solve the problem.

Complete Step By Step Answer:
Here we have been given a retardation which actually is a negative acceleration of the body in motion.
$a = - kV$
As we know the acceleration is a change in velocity with respect to time.
We can write it in the form
$a = \dfrac{{dV}}{{dt}}$
If we rewrite the retardation equation given in the question we get,
$\dfrac{{dV}}{{dt}} = - kV$
For further solving the problem as discussed above the integration method is to be used.
First separating out the like variables in the either sides of the equation
$\dfrac{{dV}}{V} = - k\,dt$
As given in the question initial velocity is ‘u’ and let us assume final velocity to be ‘v’. In changing initial velocity from ‘u’ to ‘v’ the time changes from ‘0’ to ‘t’
$\int\limits_u^v {\dfrac{{dV}}{V} = - k\int\limits_0^t {dt} }$
Integration
$\int {\dfrac{{dx}}{x} = \ln x\,\,and\int {dt = t} }$
Solving the integration we get
$\left[ {\ln V} \right]_u^v = - k\left[ t \right]_0^t \\\ put\,\lim its \\\ \Rightarrow \left[ {\ln v - \ln u} \right] = - kt$
As we know from the rules of logarithms
$\ln a - \ln b = \ln \dfrac{a}{b}$
$\Rightarrow \ln \dfrac{v}{u} = - kt \\\ \Rightarrow \dfrac{v}{u} = {e^{ - kt}}$
By further solving we get
$v = u{e^{ - kt}}...(i)$
As we know acceleration can also be written in terms of displacement as
$a = V\dfrac{{dV}}{{dx}} \\\ \Rightarrow - kV = V\dfrac{{dV}}{{dx}}$
Separating the like variables and integrating we get
$- k\int\limits_0^x {dx} = \int\limits_u^v {dV}$
Integration
$\int {dt = t}$
Solving the integration we get
$- k\left[ x \right]_0^x = \left[ {\;V} \right]_u^v \\\ put\,\lim its \\\ \Rightarrow - kx = v - u$
Putting the value of (i) in the equation we get,
$kx = u - u{e^{ - kt}} \\\ \Rightarrow x = \dfrac{u}{k}(1 - {e^{ - kt}})$
Hence the distance covered in t seconds is
$x = \dfrac{u}{k}(1 - {e^{ - kt}})$
The correct answer is option (A).

Note :
The motion of an object in a straight line is also called a rectilinear motion. We can obtain some simple equations for the rectilinear motion with the uniform acceleration. Basically they are based on Newton's laws of motion. The study of such motion is also called kinematics.