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 ?
Answers
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:
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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