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What is displacement in nth second and how to calculate it?
Asked by ayan1.chatterjee | 24 Jan, 2021, 07:08: PM
Equation of motion that relates the displacement S , intial speed u , uniform acceleration a and time t  as

S = ( u × t ) + [ (1/2) × a × t2 ]  ....................(1)

nth second displacement ΔS is difference of displacement Sn upto tn and displacement Sn-1 upto tn-1

ΔS = Sn - Sn-1 ...................(2)

Sn =  ( u × tn ) + [ (1/2) × a × tn2 ]

Sn-1 =  ( u × tn-1 ) + [ (1/2) × a × tn-12 ]

ΔS = u × ( tn - tn-1 ) + [ (1/2) × a × ( tn2 - tn-12 ) ]  ......................(5)

( tn - tn-1 ) = 1 s  ;

( tn2 - tn-12 ) = ( tn + tn-1 )( tn - tn-1 )

Hence Eqn.(5) becomes ,     ΔS = u  + [ (1/2) × a × ( tn + tn-1 ) ]  ......................(6)

Example :-

Let initial speed  u = 10 m/s  and acceleration a = 1 m/s2

Displacement at 10th second ,  ΔS = 10 + { (1/2) × 1 × (10+9) } = 19.5 m
Answered by Thiyagarajan K | 24 Jan, 2021, 08:07: PM

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