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 = 1 at the point x = 1 No other information is available about its motion at intermediate time `(0 lt t lt 1)`. If `alpha` denotes the instantaneous acceleration of the particle, then

Explanation

Since the body is at rest at `x = 0` and `x = 1`. Hence `alpha` cannot be positive for all time in the interval `0 le t le 1`.

Total time, `t = 1s`. And total displacement, `s = 1m`
`s =` area under `v - t` graph
`:.` height or `v_("max") = (2s)/(t) = 2m//s` is fixed `("area or "s = (1)/(2) xx t xx v_("max"))`
In case 1 : acceleration `gt 4 m//s^(2)` while retardation `lt 4 m//s^(2)` and
In case 2: acceleration = retardation `= 4 m//s^(2)`
In case 3: acceleration `lt 4 m//s^(2)` and retardation `gt 4 m//s^(2)`
Hence `|alpha| ge 4 m//s^(2)` at some point or points on its path.