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For the one dimensional motion, describe...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
A, D
Given, `x = t - sint`
velocity `v = (dx)/(dt) = (d)/(dt) [t - sin t]`
`= 1- "cost"`
Acceleration `a = (dv)/(dt) = (d)/(dt) [1-"cost"] = sint`
As acceleration `a gt 0` for all `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) = - sint`
When , `t = 0, x = 0 , x = +1 , a = 0`
When, `t = (pi)/(2), x = 1, v = 0, a = - 1`
When `t = pi, x = 0, x = -1, a = 1`
When `t = 2pi, x = 0, x = 0, a = 0`
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