0
$\begingroup$

A question in the third chapter of Exemplar Problems in Physics for Class XI goes thus:

(3.9) For the one-dimensional motion described 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 $2$.

The answers are (a) and (d).

I do not quite understand how they've arrived at the answers despite taking the first- and second-order derivatives of $x(t)$. A detailed explanation as to why the two options are correct would be helpful.

$\endgroup$

1 Answer 1

1
$\begingroup$

$$v(t) = x'(t) = 1 - cos \;t$$

Since $-cos \;t$ lies between -1 and 1, then $1 - cos \;t$ lies between 0 and 2. Therefore $v(t) = x'(t) \geq 0$. Note that beacuse $x(0) = 0$ and $x'(t) > 0$ in $(0, 2\pi)$, by a corollary of Lagrange Theorem we conclude that $x(t) > 0$ in $(0, 2\pi)$. Thus, $x(t) > 0$ for all $t > 0$ beacuse the derivative is non-negative.

Clearly (b) is false and (c) is false beacuse $a(t) = v'(t) = sin \;t$, which can be any value in $[-1, 1]$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.