Question

Given, $Nk{{g}^{-1}}$ is the unit of:
A) retardation
B) acceleration
C) rate of change of velocity
D) all the above

Answer 1

Hint: Force is defined as the product of mass and acceleration. Retardation is also called negative acceleration. Acceleration is also called a rate of change of velocity.$N$ is the SI unit of force and $kg$ is the SI unit of mass.

Complete answer:
We know that force acting on an object is defined as the product of mass of the object and the acceleration of the object. Mathematically, force is given by
$F=ma$
where
$F$ is the force acting on an object
$m$ is the mass of the object
$a$ is the acceleration of the object
Let this be equation 1.
The SI unit of force is newton$(N)$ while the SI unit of mass is kilogram$(kg)$. $1N$ is defined as the force acting on an object of mass $1kg$, to accelerate the object by $1m{{s}^{2}}$. Clearly, to accelerate a mass of $1kg$by $1m{{s}^{-2}}$, a force of $1N$ is required. Therefore,
$1m{{s}^{-2}}=\dfrac{1N}{1kg}\Rightarrow m{{s}^{-2}}=Nk{{g}^{-1}}$
where
$m{{s}^{-2}}=Nk{{g}^{-1}}$ is the SI unit of acceleration
Now, we know that deceleration is nothing but negative acceleration. Therefore, deceleration also has the same unit as that of acceleration. In the case of deceleration, only the direction of motion of the object is changing, but not its unit. An example of deceleration is the negative acceleration caused while braking cars and buses.
We also know that acceleration is defined as the rate of change of velocity. Suppose the velocity of a body is changing with respect to time, we call the rate of change of velocity of the body as the acceleration of the body. Mathematically, acceleration is given by
$a=\dfrac{dv}{dt}$
where
$a$ is the acceleration of a body
$dv$ is the change in velocity of the body in a time interval $dt$
Let this be equation 2.
Clearly, a change in velocity of $1m{{s}^{-1}}$ in $1s$ causes an acceleration of $1m{{s}^{-2}}$. Therefore, unit of rate of change of velocity is also equal to $Nk{{g}^{-1}}$

Hence, the correct answer is option $D$.

Note:
Acceleration is also defined as the second differential of the position of a body. We know that velocity of a body is equal to the change in position of the body with respect to time.
$v=\dfrac{dx}{dt}$
Also, acceleration of the body is equal to the change in velocity of the body with respect to time.
$a=\dfrac{dv}{dt}$
Using the expression for velocity in the above expression for acceleration, we have
$a=\dfrac{d}{dt}\left( \dfrac{dx}{dt} \right)=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$
Thus, acceleration is also defined as the second differential of the position of a body.