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For the one-dimensional motion, describe by $x = t - \sin t$

$a)\; x(t)>0 for \; all\; t>0$

$b) v(t)>0 for\; all\; t>0$

$c) a(t)>0\; for \; all\; t>0$

$d) v(t) lies\; between\; 0 \; and \; 2$

Answers (1)

The correct answer is the option:

$a)\; x(t)>0 for \; all\; t>0$

$d) v(t) lies\; between\; 0 \; and \; 2$

Explanation: Now, $x = t -\sin t$

We know that $v=\frac{dx}{dt}$

$=1-\cos \; t$

Now, $a=\frac{dv}{dt}$

$=\frac{d(1-\cos \; t)}{dt}$

$=+\sin\; t$

v_max will be,

$V_{max}=1-(-1)$

$= 1+1 = 2$

. v_min will be,

$v_{min} = 1-1$

$=0$

Thus, it is clear that v lies between 0 to 2 and option (d) is verified.

Now, $x = t -\sin t$

Thus, sin t lies between 1 and -1 for all t > 0.

Thus, x will always be positive and option (a) is verified.

Now, $v = 1 -\cos \; t$

$v= 0, when \; t = 0$

$v= 1, when\; t = \frac{\pi }{2}$

$v=2, when t = \pi$

& $v= 0, when\; t = 2\pi .$

Hereby opt (b) is discarded.

Now, $a = \sin \; \; t$

$a = 0, when\; t = 0$

$a = 1, when \; t = \frac{\pi }{2}$

$a = 0, when\; t = \pi$

& $a = 1, when\; t = 2\pi .$

Thus, acceleration can be negative as well, and hence opt (c) is also discarded here.

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