Question

A particle experiences constant acceleration for 20 seconds after starting from rest. It is travels a distance S1 in the first 10 seconds and a distance S2 in the next 10 second.. then

Answer 1

Method = 1

Let a be the constant acceleration of the

particles

then,

S= ut + 1 / 2 at^2

S1 = 0 + 1 / 2 a (10) ^2

S1 = 50a ..... (1)

S2 = 0 + 1 / 2 a (20) ^2

S2 = 200a ... (2)

Distance travel from 10s to 20s

200a - 50a = 150a

S1 = S2

50a = 150a

S2 = 3S1 = ANSWER

METHOD 2

Let a be the constant acceleration then,

S = ut + 1 / 2 at^2

S1 = 0 + 1 / 2 a (10) ^2

S1 = 50a

Velocity after 10 seconds is:-

v = u + at

v = 0 + 10a

So,

S2 = 10a X 10 + 1 / 2 X a X 100

S2 = 150a

S1 = S2 = 50a = 150a

S2 = 3S1

Answer 2

let constant acceleration is a

distance travels in 20 seconds 

s = 0*20+a*20²/2

s = 200a --------

let constant acceleration is a

distance travels in 20 seconds 

s = 0*20+a*20²/2

s = 200a -------------(1)

distance travels in first 10 seconds 

s1 = 0*10+a*10²/2

s1 = 50a -----------------(2)

distance travels in next 10 seconds 

s2 = s-s1

s2 = 200a-50a = 150a ----------------(3)

s2/s1 = 150a/50a=3

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