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Question: A particle with acceleration \(a = - kv\)(where k is the constant and v is velocity) crosses origin ...

A particle with acceleration a=−kva = - kv(where k is the constant and v is velocity) crosses origin with velocityvoi^{v_o}\hat i. Velocity of the particle at point(L,O)\left( {L,O} \right) is:
A. vo−kL{v_o} - kL
B. vo+kL{v_o} + kL
C. vo2−kL2\sqrt {{v_o}^2 - k{L^2}}
D. vo2+kL2\sqrt {{v_o}^2 + k{L^2}}

Explanation

Solution

Hint:- The acceleration of the particle is given a=−kva = - kv here we can observe that the motion of body and the direction of the acceleration is opposite to each other which means that after sometime the body will come to rest and then start to move in the opposite direction compared to the initial direction.
Formula used: The formula of the acceleration is given by,
a=(dvdx)va = \left( {\dfrac{{dv}}{{dx}}} \right)v
Where v is the velocity and a is the acceleration of the particle.

Complete step-by-step solution
It is given in the problem that a particle is moving with the acceleration a=−kva = - kv where k is the constant and v is the velocity and the particle crosses the origin with the velocity of voi^{v_o}\hat i and we need to find the value of velocity of the particle at point(L,O)\left( {L,O} \right).
As the acceleration is given as,
a=(dvdx)va = \left( {\dfrac{{dv}}{{dx}}} \right)v
Where v is the velocity and a is the acceleration of the particle.
As it is given that the velocity of the particle at origin is equal to voi^{v_o}\hat i and the velocity of the particle is asked at the point(L,O)\left( {L,O} \right).
⇒a=(dvdx)v\Rightarrow a = \left( {\dfrac{{dv}}{{dx}}} \right)v………eq. (1)
Also
a=−kva = - kv………eq. (2)
Now equating equation (1) and (2) we get,
⇒(dvdx)v=−kv\Rightarrow \left( {\dfrac{{dv}}{{dx}}} \right)v = - kv
Further solving we get,
⇒(dvdx)=−k\Rightarrow \left( {\dfrac{{dv}}{{dx}}} \right) = - k
⇒dv=−kdx\Rightarrow dv = - kdx
Now integrating the above relation for length 00 to LL and velocity from vo{v_o} to vv we get,
⇒∫vovdv=−∫0Lkdx\Rightarrow \int\limits_{{v_o}}^v {dv} = - \int\limits_0^L {kdx}
⇒v−vo=−∫0Lkdx\Rightarrow v - {v_o} = - \int\limits_0^L {kdx}
⇒v−vo=−k∫0Ldx\Rightarrow v - {v_o} = - k\int\limits_0^L {dx}
⇒v−vo=−k(L−0)\Rightarrow v - {v_o} = - k\left( {L - 0} \right)
⇒v−vo=−kL\Rightarrow v - {v_o} = - kL
⇒v=vo−kL\Rightarrow v = {v_o} - kL
So the speed of the particle at point (L,O)\left( {L,O} \right) is equal to v=vo−kLv = {v_o} - kL. The correct option for this problem is option A.

Note:-
The direction of the velocity is given in the increasing x-axis which is represented byi^\hat i. It is advisable for the students to learn and understand the basic operations of the integration and differentiation as it can be used to solve these kinds of problems.