A particle starts moving from the position of rest under a constant acc. If it travels a distance x in t sec, what distance will it travel in next t sec ?

Explanation

As acc. is constt.,from 2nd equation of motion, i.e., `s = ut + (1)/(2) at^(2)` we have
`x = (1)/(2) at^(2)` [as `u = 0`]....(i)
Now if it travels a distance `y` in next t sec., the total distance travelled in `(t + t = 2t)` sec will be `x + y`, so
`x + Y = (1)/(2) a (2t)^(2)` ....(ii)
Dividing Eqn. (ii) by (i),
`(x + y)/(x) = 4` or `y = 3x`
Alternative solution: From 2nd equation of motion, we have
`x = (1)/(2) at^(2)`
The velocity of particle after time t from 1 st equation of motion will be
`v = 0 + a t`, i.e., `v = at`
Now for next `t` sec it will be the initital velocity, so again from 2nd Eqn. of motion,
i.e., `s = u t + (1)/(2) at ^(2)`
we have, `y = (at) t + (1)/(2) at^(2) = (3)/(2) at^(2)`.....(iv)
Dividing eqn. (iv) by (iii),
`(y//x) = 3`, i.e., `y = 3x`