Question

a body cover 12 m in 2 ND second and 20m in 4th second find what distance the body will cover in 4 second after the 5th second

Answer 1

Increase in distance per second is (20 – 12)/ 2 = 4 m/s

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Distance covered during 2nd second = 12= 3*4

Distance covered during 4th second =20=5*4

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Distance covered during 6th second =28=7*4 ( after 5 seconds )

Distance covered during 9th second =40 = 10*4

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Distance covered in 4 second after 5 th second is

4 times the average of 28 and 40

(28+40)*4 / 2=136 m

For example:

Given that the motion is in 1 Dimension and the body doesn't change it's direction (i.e. velocity is in the direction of acceleration), we have:

s=u+a(2t−1)/2s=u+a(2t−1)/2

12 = u + a(2*2 - 1)/2
=> 12 = u + 3a/2
=> 2u + 3a = 24

and 

20 = u + a(2*4 - 1)/2
20 = u + 7a/2
=> 2u + 7a = 40

From 1 and 2:
a = 4
u = 6

Now, distance covered in 4s after 5th second = Distance covered in 9 seconds - distance covered in 5 seconds

= 6*9 + 1/2 * 4 * 9 * 9 - (6 * 5 + 1/2 * 4 * 5 * 5)
= 54 + 162 - (30 + 50)
= 136 m