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Question: Given, \(Nk{{g}^{-1}}\) is the unit of: A) retardation B) acceleration C) rate of change of ve...

Given, Nkg1Nk{{g}^{-1}} is the unit of:
A) retardation
B) acceleration
C) rate of change of velocity
D) all the above

Explanation

Solution

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.NN is the SI unit of force and kgkg 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=maF=ma
where
FF is the force acting on an object
mm is the mass of the object
aa is the acceleration of the object
Let this be equation 1.
The SI unit of force is newton(N)(N) while the SI unit of mass is kilogram(kg)(kg). 1N1N is defined as the force acting on an object of mass 1kg1kg, to accelerate the object by 1ms21m{{s}^{2}}. Clearly, to accelerate a mass of 1kg1kgby 1ms21m{{s}^{-2}}, a force of 1N1N is required. Therefore,
1ms2=1N1kgms2=Nkg11m{{s}^{-2}}=\dfrac{1N}{1kg}\Rightarrow m{{s}^{-2}}=Nk{{g}^{-1}}
where
ms2=Nkg1m{{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=dvdta=\dfrac{dv}{dt}
where
aa is the acceleration of a body
dvdv is the change in velocity of the body in a time interval dtdt
Let this be equation 2.
Clearly, a change in velocity of 1ms11m{{s}^{-1}} in 1s1s causes an acceleration of 1ms21m{{s}^{-2}}. Therefore, unit of rate of change of velocity is also equal to Nkg1Nk{{g}^{-1}}

Hence, the correct answer is option DD.

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=dxdtv=\dfrac{dx}{dt}
Also, acceleration of the body is equal to the change in velocity of the body with respect to time.
a=dvdta=\dfrac{dv}{dt}
Using the expression for velocity in the above expression for acceleration, we have
a=ddt(dxdt)=d2xdt2a=\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.