Question

For the one dimensional motion described by $x=t-\cos t(t>0)$ where $x$ is position and $t$ is time. Which of the following options is incorrect?

• A. $0\le v\le 2$ for all $t>0$
• B. $0\le a\le 1$ for all $t>0$
• C. $-1\le a\le 1$ for all $t>0$
• D. At $t-0,x=-1$ and $v=+1$

Answer 1

The correct option is C $-1\le a\le 1$ for all $t>0$

$x=t-\cos t-1\le \sin t\le 1$

$v=\frac{dx}{dt}=1+\sin t-1\le \cos t\le 1$

$a=\frac{d^{2}x}{d{t}^2}=\cos t0\le 1+\sin t\le 2$

$-1\le a\le 1,0\le v\le 2$