Question

A body covers 12 m in 2nd second and 20 m
in 4th second. How much distance will it cover in 4 seconds
after the 5th second ?​

Answer 1

Answer:

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:

40m

Explanation:

WE KNOW THAT,

first term=a+d

second term=a+2d

so,

12=a+d. -------------------[1]

and,

20=a+3d. --------------------[2]

From [1] ,

a=12-d. -----------------3

From 2 and 3,

20=12-d+3d

20-12=2d

8= 2d

d=8/2

d=4

substituting the value of d in 1

we get,

12=a+4

a=12-4

a=8

therefore in the 9th second it will be a+8d,

a+8d=8+8×4

=8+32

=40

therefore in the 9th second the body covers 40m.

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