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For the one dimensional motion, describe...

For the one dimensional motion, described by `x=t-sint`

A

`x(t)gt0` for all `tgt0`

B

`v(t)gt0` for all `tgt0`

C

`a(t)gt0` for all `tgt0`

D

`v(t)` lies between 0 and 2

Text Solution

Verified by Experts

The correct Answer is:
A, D

`x=t-sint`
velocity `v=(dx)/(dt)=(d)/(dt)[t-sint]`
`=1-cost`
Acceleration `a=(dv)/(dt)=(d)/(dt)[1-cost]=sint`
As acceleration `agt0` for all `tgt0`
Hence, `x(t)gt0` for all `tgt0`
velocity `v=1-cost`
When `cost=1` velocity `v=0`
`V_(max)=1-(cost)_(min)=1-(-1)=2`
`V_(min)=1-(cost)_(max)=1-1=0`
Hence, v lies between 0 and 2.
Acceleration `a=(dv)/(dt)=-sint`
When `t=0,x=0,x=+1,a=0`
When `t=(pi)/(2),x=1,v=0,a=-1`
`t=pi,x=0,x=-1,a=1`
When `t=2pi,x=0,x=0,a=0`
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