A body moving with uniform acceleration …
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Yogita Ingle 4 years, 4 months ago
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
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