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

Options

A. `x (t) gt 0` for all `t gt 0`
B. `v (t) gt 0` for all` t gt 0`
C. `a (t) gt 0` for all `t gt 0`
D. v (t) lies between 0 and 2

Answer

A, B

Explanation

Given, `x =t - sin t `
velocity `v = (dx)/(dt) = (d)/(dt) [t - sin t]`
`=1 - cos t`
Acceleration `a = (dv)/(dt) = (d)/(dt) [1- cos t] sin t `
As acceleration `a gt 0 ` for all `t gt 0 (because sin t gt 0)`
Hence,` x (t) gt 0` for all `t gt 0`
Velocity `v =1 - cos t`
When, `cos t = 1,` velocity `v =0`
`V_(max) =1 - (cos t ) _(min) =1 - (-1) =2`
`V _(min) =1- (cos t) _(max) 1-1 =0`
Hence, v lies between 0 and 2.
Acceleration `a - (dv)/(dt) =- sin t`
When `t =0, x=0, v =+ 1, a =0`
When `t = 90^(@),x = 1, v =0, a =-1`
When `t = 180^(@) , x=0, v =-1, a =1`
When `t = 360 ^(@),x =0 , v =0 ,a=0`