Question

The initial velocity of a particle is $ u $ and the acceleration is given by $ \left( {kt} \right) $ , where $ k $ is a positive constant. The distance traveled in time $ t $ is:
(A) $ s = u{t^2} - k{t^2} $
(B) $ s = ut - \left( {\dfrac{{k{t^3}}}{6}} \right) $
(C) $ s = ut + \left( {\dfrac{{k{t^3}}}{2}} \right) $
(D) $ s = \left( {\dfrac{{u{t^2}}}{2}} \right) + \left( {\dfrac{{k{t^3}}}{6}} \right) $

Answer 1

Hint: We are given with the initial velocity and the acceleration of the particle and are asked to evaluate the distance traveled in a given time. Thus, we will use the appropriate equation of motion and then put in the values of the initial velocity and acceleration and then evaluate the final answer.

Formulae Used:
 $ s = ut + \dfrac{1}{2}a{t^2} $
Where, $ s $ is the distance traveled by the particle, $ u $ is its initial velocity, $ a $ is its acceleration and $ t $ is the time of travel.

Complete Step By Step Solution
Here Given,Initial velocity of the particle $ = u $
Acceleration of the particle $ = kt $
Now, Applying the relevant equation of motion, we get
 $ s = ut + \dfrac{1}{2}(kt){t^2} $
Further, we get
 $ s = ut + \dfrac{1}{2}kt \times {t^2} $
Then, we get
 $ s = ut + \left( {\dfrac{{k{t^3}}}{2}} \right) $
Hence, the correct answer is (C).

Note
The answer we got clearly implies that as time proceeds, the distance also increases. This is because of the given that $ k $ is a positive constant. Distance and displacement are different quantities. Distance increases as the body moves. But Displacement is calculated between starting point and end point.