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Question: The distance travelled by a body in \[\mathop 4\nolimits^{th} \] second is twice the distance travel...

The distance travelled by a body in 4th\mathop 4\nolimits^{th} second is twice the distance travelled in 2nd\mathop 2\nolimits^{nd} second. If the acceleration of the body is 3m/s23m/s^2, then its initial velocity is?

Explanation

Solution

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+12at2d = ut + \dfrac{1}{2}a{t^2} ….(1)
Where uu is the initial velocity, aa is the acceleration of the body, and tt is the time.

Complete step by step answer:
Step 1: It is given in the question that d4=2d2\mathop d\nolimits_4 = 2\mathop d\nolimits_2 and a = 3$$$m/s^2$. Let us consider the initial velocity is u.Usingequation(1). Using equation (1) \mathop d\nolimits_4 andand\mathop d\nolimits_2 canbecalculatedasgivenbelow.can be calculated as given below. \mathop d\nolimits_4 = distancetravelledbythebodyin4secondsdistancetravelledbythebodyin3secondsSo,distance travelled by the body in 4 seconds – distance travelled by the body in 3 seconds So, \mathop d\nolimits_4 = u(4) + \dfrac{1}{2}a{(4)^2} - \left\{ {u(3) + \dfrac{1}{2}a{{(3)}^2}} \right\} \mathop d\nolimits_4 = u + \dfrac{7}{2}a.....(2)And,..... (2) And, \mathop d\nolimits_2 = distancetravelledbythebodyin2secondsdistancetravelledbythebodyin1secondSo,distance travelled by the body in 2 seconds – distance travelled by the body in 1 second So,\mathop d\nolimits_2 = u(2) + \dfrac{1}{2}a{(2)^2} - \left\{ {u(1) + \dfrac{1}{2}a{{(1)}^2}} \right\} \mathop d\nolimits_2 = u + \dfrac{3}{2}a$$..... (3)

Step 2: Now from the given condition i.e. d4=2d2\mathop d\nolimits_4 = 2\mathop d\nolimits_2 …...(4)
From keeping the values of d4\mathop d\nolimits_4 and d2\mathop d\nolimits_2 from equations (2) and (3) respectively in equation (4), we will get-
u+72a=2(u+32a)=2u+62au + \dfrac{7}{2}a = 2\left( {u + \dfrac{3}{2}a} \right) = 2u + \dfrac{6}{2}a
u+72a=2u+62au + \dfrac{7}{2}a = 2u + \dfrac{6}{2}a
2uu=72a62a2u - u = \dfrac{7}{2}a - \dfrac{6}{2}a
u=a2u = \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=32m/su = \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.