Question

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

• 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. all of these

Answer 1

The correct option is A x(t) > 0 for all t > 0

The position of the particle is given as a function of time.

$x=t-\sin t$

The velocity can be obtained by differentiating the given expression.

$v=\frac{dx}{dt}=\frac{d}{dt}[t-\sin t]=1-\cos t$

The acceleration can be obtained by differentiating the expression of velocity w.r.t. time

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

$a=\frac{d}{dt}[1-\cos t]=\sin t$

As acceleration $a>0$ for all $t>0$

Hence, $x\left(t\right)>0$ for all $t>0$

velocity $v=1-\cos t$

if, $\cos t=1,$

the velocity will be $v=0$

$v_{max}=1-(\cos t{)}_{min}=1-(-1)=2$

$v_{min}=1-(\cos t{)}_{max}=1-1=0$

Hence, v lies between $0$ and $2$.

For acceleration

$a=\frac{dv}{dt}=-\sin t$

When $t=0;x=0,v=0,a=0$

When $t=\frac{\pi }{2};x=$ positive , $v=0,a=-1$ (negative)

When $t=\pi ,x=$ positive, $v=$ positive , $a=0$

When $t=2\pi ,x=0,v=0,a=0$