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 times `( 0 lt t lt 1)` . If `alpha ` denotes the instantaneous accelartion of the particle , then :

Options

A. ` alpha ` cannot remain positive for all t in interval ` 0le t le 1`
B. `|alpha| `must be `ge` 4 at same point or points 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

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 lt t lt 1` . Therefore, first the particle is accelerated and then retarded. Now, total time t = 1s (given) Total displacement s = 1m (given) S = area under v – t graph
Height or ` v_(max) =(2s)/(t) = 2m//s ` is also fixed
[Area or ` s=(1)/(2) xx txx v _(max) ]`
In case 2 : Acceleration = Retardation = ` 4 m//s^(2) `
In case 1: Acceleration `gt 4m//s^(2) and `Retardation `lt 4 m//s^(2) `
while in case 3: Acceleration `lt 4m//s^(2) ` and Retardation Hence , `|alpha |ge 4 ` at some point or points in its path.
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