A particle of mass `m` moves on the ` x- axis ` as follows : it starts from rest at ` t = 0`, from the point `x = 0`, and comes to rest at ` t = l` at the point `x = 1`. No other information is available about its motion at intermediate times `( 0 lt t lt l)` . If `alpha ` denotes the instantaneous accelartion of the particle , then :

Options

A. `alpha ` cannot remain positive for all `t` in the interval ` 0 le t le 1 `.
B. `|alpha |` cannot exceed `2` at any point in its path .
C. `| alpha |` must be ` ge 4` at some point or points in its path .
D. ` alpha` must change sign during the motion , but no other assertion can be made with the information given .

Answer

A, C, D

Explanation

Note : `alpha ` cannot remain positive for all `t` in the interval ` 0le tle 1`. This is because since the body starts from rest , it will first accelerate . Finally it stops therfore ` alpha` will become negative . Therfore `alpha` will change its direction. options `(a)` and `(d)` are correct .
Let the particle accelerate uniformly till half the distance `(A to B)` and then retard uniformly in the remaining half distance `(B to C)` .
The total time is `1 sec`. Therfore the time taken from ` A to B is 0.5 sec `.
For `A to B`
` S = ut + 1/2 at^(2) 0.5 = 0 + 1/2 xx a xx (0.5)^(2)` ltbr gt `:. a = 4 m//s^(2) `
`:. V_(B) = 0 + 4 xx 0.5 = 2 m//s^(2) `
Note : Now , if the particle accelerates till ` B_(2) ` then for covering the same total distance in same time , acceleration should be less than ` 4 m //s^(2) ` but `| decceleration|` till ` B_(1)` , then for covering the same total distance in the same time , the acceleration should be greater than ` 4 m//s^(2) ` and `|deceleration | lt 4 m//s^(2) `. The same is dipcted by the graph .
So, the `|acceleration |` must be greater than or equal to ` 4 m//s^(2) ` at some point or points in the path .