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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+a2(2n1){S_{nth}} = u + \dfrac{a}{2}\left( {2n - 1} \right), where uu is the initial velocity, aa is the acceleration and Snth{S_{nth}} is the distance travelled by the body in nthnth second.

Explanation

Solution

First consider the displacements for nsecn\sec and (n1)sec\left( {n - 1} \right)\sec . Now the equation can be obtained by integrating the kinematic equation v=u+atv = u + at (where v=dsdtv = \dfrac{{ds}}{{dt}}) and then apply the required limit. Here time is considered as a discrete number (i.e., 11sec, 22sec, 33sec, … , nn sec).

Complete step by step answer:
Let a body start from O to P1 in (n1)\left( {n - 1} \right) second and to P2 in nn second as shown in the following figure. Then, P1P2{P_1}{P_2} is the distance travelled by the body in nthnth second.
Snth=SnSn1{S_{nth}} = {S_n} - {S_{n - 1}}
Where Sn{S_n} and Sn1{S_{n - 1}} are the distances covered by the body in nn and n1n - 1 second respectively.

If ds\overrightarrow {ds} be the small displacement moved by the body in a small time dtdt, instantaneous velocity v\vec v of the body is
v=dsdt\vec v = \dfrac{{\vec ds}}{{dt}}
ds=vdt\Rightarrow ds = vdt
We know that v=u+atv = u + at
Where uu and aa are the initial velocity and acceleration of the body respectively.
ds=(u+at)dt\Rightarrow ds = \left( {u + at} \right)dt

Integrating the above equation between the region P1P2 .
Sn1Snds=n1n(u+at)dt\int_{{S_{n - 1}}}^{{S_n}} {ds} = \int_{n - 1}^n {\left( {u + at} \right)dt}
On further simplification,
[S]Sn1Sn=u[t]n1n+a[t22]n1n\left[ S \right]_{{S_{n - 1}}}^{{S_n}} = u\left[ t \right]_{n - 1}^n + a\left[ {\dfrac{{{t^2}}}{2}} \right]_{n - 1}^n
Substitute the limits and calculate
SnSn1=u[n(n1)]+a[n2(n1)22]\Rightarrow {S_n} - {S_{n - 1}} = u\left[ {n - \left( {n - 1} \right)} \right] + a\left[ {\dfrac{{{n^2} - {{\left( {n - 1} \right)}^2}}}{2}} \right]
Simplify further,
SnSn1=u+a2(2n1)\Rightarrow {S_n} - {S_{n - 1}} = u + \dfrac{a}{2}\left( {2n - 1} \right)
Since, SnSn1=Snth{S_n} - {S_{n - 1}} = {S_{nth}}

Therefore, Snth=u+a2(2n1){S_{nth}} = u + \dfrac{a}{2}\left( {2n - 1} \right) is the correct relation.

Note: While doing the integration, keep uu and aa constant. If a body starts from rest, u=0u = 0.Therefore, displacement in nth second is Snth=a2(2n1){S_{nth}} = \dfrac{a}{2}\left( {2n - 1} \right). It should be kept in mind that the equations of kinematics are valid only for uniformly accelerated motion i.e., when a=a = constant.