Question

A body moving with uniform accleration cover 20m in 2nd second and 30m in 4th second of its motion the distance moved by it in 6th second is

Answer 1

Answer:

40m

Explanation:

Given:

distance covered in 2nd second = 20 m.

distance covered in 4th second = 30 m.

We know =>

Snth = u + a(2n-1)/2

S2nd = u+a(2×2-1)

20 = u+3a/2

20 = (2u+3a)/2

40 = 24 + 3a _____(1)

again

S4th = u +a(2×4-1)/2

30 = u+7a/2

30=(2u+7a)/2

60 = 2u+7a _____(2)

(1)-(2)

2u+3a-2u-7a = 40 - 60

-4a=-2a

a=5

putting the value of a in equation 1, we get

2u+ 3(5)=40

2u=40-15

u=25/2

NOW —>

S6th = u+a(2n-1)

S6th = 25/2 + 5(2×6-1)/2

" = 25/2 + 55/2

" = 80/2

" = 40

Hence,distance covered in the 6th second = 40 m

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