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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)
Since , x =t - sin t .
v = 1 - cot t
and a = sin t .
`x (t) gt 0 ` for all `t gt 0`
At `t = 0 , v(t) =0.` So , v(t) can be zero for one value of t.
At t = 0 and `pi` ,a (t) = 0 .
If `t = 0`,v (t) = 0 , if `t = pi , v(t) = 1-(-1) = 2`
So , v(t) lies between 0 and 2 .
Here we can see that v(t) is also equal to zero at certain instants and hence is not always greater then zero . So option (b) is not correct .
The correct options are (a) and (d).
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