Question
Question: Check the correctness of the relation \({S_{nth}} = u + \dfrac{a}{2}\left( {2n - 1} \right)\), where...
Check the correctness of the relation Snth=u+2a(2n−1), where u is the initial velocity, a is the acceleration and Snth is the distance travelled by the body in nth second.
Solution
First consider the displacements for nsec and (n−1)sec. Now the equation can be obtained by integrating the kinematic equation v=u+at (where v=dtds) and then apply the required limit. Here time is considered as a discrete number (i.e., 1sec, 2sec, 3sec, … , n sec).
Complete step by step answer:
Let a body start from O to P1 in (n−1) second and to P2 in n second as shown in the following figure. Then, P1P2 is the distance travelled by the body in nth second.
Snth=Sn−Sn−1
Where Sn and Sn−1 are the distances covered by the body in n and n−1 second respectively.
If ds be the small displacement moved by the body in a small time dt, instantaneous velocity v of the body is
v=dtds
⇒ds=vdt
We know that v=u+at
Where u and a are the initial velocity and acceleration of the body respectively.
⇒ds=(u+at)dt
Integrating the above equation between the region P1P2 .
∫Sn−1Snds=∫n−1n(u+at)dt
On further simplification,
[S]Sn−1Sn=u[t]n−1n+a[2t2]n−1n
Substitute the limits and calculate
⇒Sn−Sn−1=u[n−(n−1)]+a[2n2−(n−1)2]
Simplify further,
⇒Sn−Sn−1=u+2a(2n−1)
Since, Sn−Sn−1=Snth
Therefore, Snth=u+2a(2n−1) is the correct relation.
Note: While doing the integration, keep u and a constant. If a body starts from rest, u=0.Therefore, displacement in nth second is Snth=2a(2n−1). It should be kept in mind that the equations of kinematics are valid only for uniformly accelerated motion i.e., when a=constant.