Question

An object moves velocity $\overrightarrow{v}$ having constant accelaration $\overrightarrow{a}$. Which of the following expressions are variable quantities ?

• A. $\frac{d|\overrightarrow{v}|}{dt}$
• B. $\left|\frac{d\overrightarrow{v}}{dt}\right|$
• C. $\frac{d\left(|\overrightarrow{v}{|}^2\right)}{dt}$
• D. $\frac{d\left(\frac{\overrightarrow{v}}{\overrightarrow{|v|}}\right)}{dt}$

Answer 1

Answer: (A), (C), (D)

$\frac{d\left(\frac{\overrightarrow{v}}{\overrightarrow{|v|}}\right)}{dt}$

let $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$ and
$\overrightarrow{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}=\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k}$
As $\overrightarrow{a}$ is constant, so its magnitude as well as direction is not changing, but $v_x,v_y$ and $v_z$ all can be varying (any combination of these if $a_x,a_y\text{or}{a}_z$ , then corresponding velocity component may be constant.)



$A)$
$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2+v_z^2}$
Hence $\frac{d|\overrightarrow{v}|}{dt}$is variable.
$\therefore$Option(a) is correct.

$B)$
By definition
$\frac{d\overrightarrow{v}}{dt}=\overrightarrow{a}$
hence $\frac{d\overrightarrow{v}}{dt}$ is constant

$\therefore$Option(b) is incorrect.

$C)$
$|\overrightarrow{v}{|}^2=v_x^2+v_y^2+v_z^2$
$\frac{d|\overrightarrow{v}{|}^2}{dt}$ is variable
$\therefore$Option(c) is correct.

$D)$
$\frac{d\left(\frac{\overrightarrow{v}}{\overrightarrow{|v|}}\right)}{dt}=\frac{d\left(\frac{v_x\hat{i}+v_y\hat{j}+v_z\hat{k}}{\sqrt{v_x^2+v_y^2+v_z^2}}\right)}{dt}$ is variable quantity.


$\therefore$Option(d) is correct.