Question

Which of the following relations are always true? $\vec v = {\text{velocity}}$ , $\vec a = {\text{acceleration}}$ , $K = \dfrac{1}{2}m{v^2} = {\text{kinetic}}\;{\text{energy}}$.
${\text{A}}{\text{. }}\dfrac{{dK}}{{dt}} = m\vec v \cdot \vec a \\\ B.{\text{ }}\dfrac{{d\left| {\vec v} \right|}}{{dt}} = \dfrac{{\vec v \cdot \vec a}}{{\left| {\vec v} \right|}} \\\ C.{\text{ }}\dfrac{{d\left| {\vec v} \right|}}{{dt}} = \left| {\dfrac{{d\vec a}}{t}} \right| \\\ D.{\text{ }}\Delta \vec a = \left| {\int\limits_{t1}^{t2} {\vec a} {\text{ }}dt} \right| \\\$

Answer 1

Hint: Check for dimensional consistency and eliminate some options. Then we can replace the magnitude of a vector with a dot product of the vector with itself. Then differentiate the resulting expression following the product rule to find the correct options.

Complete step by step answer:
Here, trying out all the options would be time consuming. So let us first eliminate the obviously wrong options by checking the dimensions.
Option c has $\dfrac{{{\text{speed}}}}{{{\text{time}}}}$ on the Left-hand-side and $\dfrac{{{\text{acceleration}}}}{{{\text{time}}}}$ on The Right-hand-side. So this equation is dimensionally inconsistent and hence wrong.
Similarly, we know that the integral $\int {\vec a\;dt}$ is done by multiplying a small time interval with an acceleration vector. So the dimensions of the right hand side is acceleration $\times$time whereas the left hand side has the dimensions of acceleration. So option b can also be removed.

Now, let us consider option B.
We know that the magnitude of a vector can be written as the dot product of a vector with itself.
${\left| {\vec v} \right|^2} = \vec v \cdot \vec v$
$\left| {\vec v} \right| = \sqrt {\vec v \cdot \vec v}$
we are asked about $\dfrac{{d\left| {\vec v} \right|}}{{dt}}$
So let us differentiate this expression.
$\dfrac{{d\left| {\vec v} \right|}}{{dt}} = \dfrac{{d\sqrt {\vec v \cdot \vec v} }}{{dt}}$ $= \dfrac{1}{{2\sqrt {\vec v \cdot \vec v} }}\dfrac{{d\left( {\vec v \cdot \vec v} \right)}}{{dt}}$
Now, we know that $\vec v \cdot \vec v$is $|v{|^2}$since it is the dot product. So let us replace $\vec v \cdot \vec v$. Also, we know from chain rule of vector differentiation that $\dfrac{{d\left( {\vec u \cdot \vec v} \right)}}{{dt}} = \vec u \cdot \dfrac{{d\vec v}}{{dt}} + \dfrac{{d\vec u}}{{dt}} \cdot \vec v$
$= \dfrac{1}{{2\sqrt {|v{|^2}} }}\left( {\vec v \cdot \dfrac{{d\vec v}}{{dt}} + \dfrac{{d\vec v}}{{dt}} \cdot \vec v} \right)$
$= \dfrac{1}{{2v}}2 \times \vec v \cdot \dfrac{{d\vec v}}{{dt}}$ where $|v|$ is equivalent to $v$
$= \dfrac{1}{v}\vec v \cdot \dfrac{{d\vec v}}{{dt}} = \dfrac{{\vec v \cdot \vec a}}{v}$
So we see that option B is correct.
Now Let us see option A. Here, $K$ is the Kinetic energy given as
$K = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m\left( {\vec v \cdot \vec v} \right)$
We are asked to check for $\dfrac{{dK}}{{dt}}$. So let us differentiate the above expression. $\dfrac{{dK}}{{dt}} = \dfrac{{d\left( {\dfrac{1}{2}m\vec v \cdot \vec v} \right)}}{{dt}} = \dfrac{m}{2}\left( {\vec v \cdot \dfrac{{d\vec v}}{{dt}} + \dfrac{{d\vec v}}{{dt}} \cdot \vec v} \right)$
$= m\vec v \cdot \dfrac{{d\vec v}}{{dt}} = m\vec v \cdot \vec a$
So we can see that options A and B are correct whereas C and D are incorrect.

Note: Options A can be intuitively analysed as well. We can see that the RHS of the equation is $m\vec a \cdot \vec v = {F_{net}} \cdot \vec v =$ power delivered to the system. We know that power delivered to a system is the rate of doing work on a system by all forces together. Also, we know from work energy theorem that the total work done by all forces combined gives the Kinetic energy of the body. So the time derivative of kinetic energy should give us the time derivative of net work done , which is the power delivered.