Question

Let v

be the velocity of a particle at time t. Then

​

Answer 1

Answer:

B, D

Explanation:

Consider a simple example, where $\vec{v}(t) = t\hat{i} \hspace{0.1cm}+ \hspace{0.1cm}t^2 \hat{j}$

$\mid\dfrac{d\vec{v}(t)}{dt}\mid \hspace{0.1 cm}= \mid\hat{i} \hspace{0.1 cm}+\hspace{0.1 cm} 2t\hat{j}\mid = \sqrt{1 \hspace{0.1 cm}+\hspace{0.1 cm}4t^2 }$

$\dfrac{d\mid\vec{v}(t)\mid}{dt} = \dfrac{d}{dt}(\sqrt{t^2 \hspace{0.1 cm}+ \hspace{0.1 cm} t^4}) = \dfrac{1}{2\sqrt{t^2 \hspace{0.1 cm}+ \hspace{0.1 cm}t^4}}(2t \hspace{0.1 cm}+ \hspace{0.1 cm}4t^3) = \dfrac{t \hspace{0.1 cm}+ \hspace{0.1 cm}2t^2 }{t\sqrt{1 \hspace{0.1 cm}+ \hspace{0.1 cm}t^2}} = \dfrac{1 \hspace{0.1 cm}+ \hspace{0.1 cm}2t}{\sqrt{1 \hspace{0.1 cm}+ \hspace{0.1 cm}t^2}}$

As clear from the above, both these quantities are not always equal.

Hence option A is ruled out/incorrect.

They are equal at t = 0, hence option B is correct.

If $\vec{v}(t) = 2\hat{i}$, then  $\dfrac{d\mid\vec{v}(t)\mid}{dt}$ is zero. Hence C is incorrect.

$\dfrac{d\mid\vec{v}(t)\mid}{dt} \neq 0 \implies \vec{v}(t) \text { is not a constant vector}$

$\implies \dfrac{d\vec{v}(t)}{dt} \text{ is a non-zero vector}$

$\implies \mid\dfrac{d\vec{v}(t)}{dt}\mid \text{is non-zero.}$

$\text{Therefore }\dfrac{d\mid\vec{v}(t)\mid}{dt} \neq 0 \text{ implies} \mid\dfrac{d\vec{v}(t)}{dt}\mid \hspace{0.1 cm}\neq 0.$

Hence, D is correct.

Therefore, B and D are the correct options.

Hope this helps!