Question

a particle starting from test with constant accelaration travels a distance x in first 2s and a distance y in next 2s then what is the relation between x and y

Answer 1

The particle covers a distance of x  in 2 seconds
x = ut + 1/2 at²
x = 0 + 1/2 at²
x = 1/2  .a. 2²
x = 2a [here a represents the acceleration ]

As the particle covers a distance of x+y in 4 seconds;
x + y = 1/2 .a .4²
x + y = 8a
put x= 2a in the above equation then u get y = 6a

consider y/x = 6a / 2a = 3
hence   y = 3x

Hope this helps...
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Answer 2

Answer:

Y =  3X

Explanation:

S = ut + 1/2 at square                               ( S1 = X    S2 = Y

X = 2u + 1/2 a4                                         t1 = 2s    t2 = 2 +2 = 4s)

X = 2u + 2a  

since, u = 0

Therefore, X = 2a...…(1)

Y = (4u + 1/2 a16) -  ( 2u + 1/2 a4)

Y = 4u + 8a - 2u - 2a

Y = 2u + 6a

since, u = 0

Therefore, Y = 6a

here a are equal

a = Y/6

X = 2a

= 2 × Y/6

X= Y/3

Y = 3X

I hope it helpful

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