Question

A body moving with uniform acceleration describes 20m in 2nd sec and 30m in 4th sec of its motion. Calculate the distance moved by it in 6th sec.

Answer 1

Answer:

40 m

Explanation:

Given,

distance covered in 2nd s = 20 m

distance covered in 4th s = 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 = 2u + 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 = -20

=>a = 5

Putting the value of a in eq (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

=>S6th = 25/2 + 5*11/2

=>S6th = 25/2 + 55/2

=>S6th = 80/2

=>S6th = 40

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