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 ?

A particle starts moving from the position of rest under a constant ac

1.	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 ?
Answer» 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`

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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 ?

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