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. a cannot remain positive for all `t` in the interval `0letle1`
B. `|a|` cannot excessd 2 at any point in itys path
C. `|a|` must be `ge4` at some point or points in its path
D. a must change sing during the motion but no other assertion can be made with the information given.

Answer

A, C, D

Explanation

Since, the body is at rest at `x=0` and `x=1`. Hence, a cannot be positive for all time in the interval `oletle1` therefore first the particle is accelerated and then retarded now total time it `=1s` given
`S="area under" v-t` graph
Height or `v_(max)=(2s)/(t)=2 m//s` is also fixed
`["area" or S=(1)/(2)xxtxxv_(max)]`
If height and base is fixed area is also fixed in case 2 : acceleration `=` retardation `=4 m//s^(2)` in case 1: acceleration `gt4 m//s^(2)` while retardation `lt 4 m//s^(2)`. Hence `a^(3)`4 at some point in its path.