Anonymous

# a body in uniform acceleration covers 75m in6th sec 115m in 11th sec . Find displacement in 16th sec?

note: it is given 6th second not 6 second. moreover give full solution not only answer

Relevance

from the 6th sec to 11th sec is 5 seconds.

Find acceleration

Xf = .5at^2

115 = .5(a)(5)^2

230 = 25a

a = 9.2 m/s^2

assume the question ask for the displacement from 6th sec to 16th sec

so from 6th sec to 16th sec is 10 seconds

Xf = .5at^2

Xf = .5(9.2)(10)^2

Xf = 460m

just in case the question ask for the displacement from 11th sec to 16th sec.

First you got to find the speed at which the body reaches at the end of 115m

Vf^2 = 2(9.2)(115) + 0^2

Vf^2 = 46 m/s

Xf = .5at^2 + Vt + Xi

Xf = .5(9.2)(5)^2 + 56(5)

Xf = 345m

• Distance traveled in the nth second is

S = U + [n-1/2] a

75 = U + 5.5 a. and =========1

115 = U + 10.5 a ==============2

2 -1 gives

40 = 5a

a = 8 m/s^2

From 1

U = 31 m/s

Displacement in 16th sec

= 31 + 15.5 * 8 = 155m.

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