Question

The distance travelled by a body in \[\mathop 4\nolimits^{th} \] second is twice the distance travelled in \[\mathop 2\nolimits^{nd} \]second. If the acceleration of the body is $3m/s^2$, then its initial velocity is?

Answer 1

Hint: This given problem is an example of uniformly accelerated motion along a straight line i.e., the acceleration of the body will be the same throughout the whole moving process. This given problem can be solved by using the second equation of motion and the equation is as given below-
Distance traveled by the body can be given by the formula-
\[d = ut + \dfrac{1}{2}a{t^2}\] ….(1)
 Where \[u\] is the initial velocity, \[a \] is the acceleration of the body, and \[t\] is the time.

Complete step by step answer:
Step 1: It is given in the question that \[\mathop d\nolimits_4 = 2\mathop d\nolimits_2 \]and \[a = 3\]$m/s^2$.
Let us consider the initial velocity is \[u\].
Using equation (1) \[\mathop d\nolimits_4 \] and \[\mathop d\nolimits_2 \] can be calculated as given below.
\[\mathop d\nolimits_4 = \]distance travelled by the body in 4 seconds – distance travelled by the body in 3 seconds
So, \[\mathop d\nolimits_4 = u(4) + \dfrac{1}{2}a{(4)^2} - \left\{ {u(3) + \dfrac{1}{2}a{{(3)}^2}} \right\}\]
 \[\mathop d\nolimits_4 = u + \dfrac{7}{2}a\]..... (2)
And, \[\mathop d\nolimits_2 = \] distance travelled by the body in 2 seconds – distance travelled by the body in 1 second
So, \[\mathop d\nolimits_2 = u(2) + \dfrac{1}{2}a{(2)^2} - \left\{ {u(1) + \dfrac{1}{2}a{{(1)}^2}} \right\}\]
 \[\mathop d\nolimits_2 = u + \dfrac{3}{2}a\]..... (3)

Step 2: Now from the given condition i.e. \[\mathop d\nolimits_4 = 2\mathop d\nolimits_2 \] …...(4)
From keeping the values of \[\mathop d\nolimits_4 \] and \[\mathop d\nolimits_2 \] from equations (2) and (3) respectively in equation (4), we will get-
\[u + \dfrac{7}{2}a = 2\left( {u + \dfrac{3}{2}a} \right) = 2u + \dfrac{6}{2}a\]
\[u + \dfrac{7}{2}a = 2u + \dfrac{6}{2}a\]
\[2u - u = \dfrac{7}{2}a - \dfrac{6}{2}a\]
\[u = \dfrac{a}{2}\]; where \[a = 3\]$m/s^2$ in this given problem
So, \[u = \dfrac{3}{2}\]m/s

$\therefore $ The initial velocity of the body is \[u = \dfrac{3}{2} m/s\]. Hence, the correct option is (A).

Note:
While solving these types of questions one should take the difference between the distance traveled in the present amount of to the distance traveled in the just previous amount of time. If it is not solved by this method then one is going in the wrong direction.