Question

A body covers 12m in 2nd second and 20 m in 4th second . how many meter will cover in 4 second after the 5th second. substitute form sn=u+a/2(2n-1).

Answer 1

Given :
Case I:
let initial velocity = u m/s
distance=s=12m
Time =t=2sec
S=u+a(2t-1)/2
12=u+(2x2-1)/2
12=u+3a/2
2u+3a=24--------------(1)

Case II:
Time=t=4sec
distance =s=20m
20=u+a(2x4-1)/2
20=u+7a/2
2u+7a=40--------(2)
from 1 and 2:
2u+3a=24
2u+7a=40
-      -     -
***********
-4a=-16

a=4m/s2

substitute the value of a in equation 1 we get :

u=6m/s 
Case iii:
Now, distance covered in 4sec after 5second=distance covered in 9th sec-distance covered in 5 sec
=(6x9+1/24x9x9)-(6x5+1/2*4*5*5)
=54+162-(30+50)
=136m

Answer 2

Answer:

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