Question
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 second is twice the distance travelled in 2ndsecond. If the acceleration of the body is 3m/s2, then its initial velocity is?
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+21at2 ….(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 d4=2d2and a = 3$$$m/s^2$.
Let us consider the initial velocity is u.Usingequation(1)\mathop d\nolimits_4 and\mathop d\nolimits_2 canbecalculatedasgivenbelow.\mathop d\nolimits_4 = distancetravelledbythebodyin4seconds–distancetravelledbythebodyin3secondsSo,\mathop d\nolimits_4 = u(4) + \dfrac{1}{2}a{(4)^2} - \left\{ {u(3) + \dfrac{1}{2}a{{(3)}^2}} \right\}\mathop d\nolimits_4 = u + \dfrac{7}{2}a.....(2)And,\mathop d\nolimits_2 = distancetravelledbythebodyin2seconds–distancetravelledbythebodyin1secondSo,\mathop d\nolimits_2 = u(2) + \dfrac{1}{2}a{(2)^2} - \left\{ {u(1) + \dfrac{1}{2}a{{(1)}^2}} \right\}\mathop d\nolimits_2 = u + \dfrac{3}{2}a$$..... (3)
Step 2: Now from the given condition i.e. d4=2d2 …...(4)
From keeping the values of d4 and d2 from equations (2) and (3) respectively in equation (4), we will get-
u+27a=2(u+23a)=2u+26a
u+27a=2u+26a
2u−u=27a−26a
u=2a; where a = 3$$$m/s^2$ in this given problem
So, u = \dfrac{3}{2}$$m/s
∴ The initial velocity of the body is u=23m/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.