Question

The unit of retardation is
(A) $m{s^{ - 1}}$
(B) $m{s^{ - 2}}$
(C) $m$
(D) $m{s^2}$

Answer 1

Hint Retardation is the type of acceleration said to happen when the magnitude of the velocity of a body is reducing with time. Acceleration is change in velocity per unit time.

Formula used: $a = \dfrac{{\Delta v}}{t}$where $a$ is acceleration (or retardation), $\Delta v$ is the change in velocity, and $t$ is time taken for the change to happen.

Complete step by step answer
Retardation is the type of acceleration whereby the magnitude of the velocity is reducing with time, given by
$\Rightarrow a = \dfrac{{\Delta v}}{t}$
Thus, its unit is the unit of acceleration.
From the formula above, the unit of acceleration is given by replacing the quantities with their unit.
Hence,
$\Rightarrow {a_u} = \dfrac{{m{s^{ - 1}}}}{s} = m{s^{ - 2}}$ where ${a_u}$ is the unit of acceleration.
$\therefore {a_u} = m{s^{ - 2}}$
Since retardation has the same unit as acceleration, then the unit of retardation is also $m{s^{ - 2}}$.
Hence, the right answer is B.

Additional Information
We know that force (measured in Newton) is given by mass of a body multiplied by the acceleration of that body. Forces allow a body to be accelerated (increase velocity) for as long as the force is applied. However, when a force acts in such a way that it retardates (or decelerates) a body (reduces velocity), the force is said to be retardation, and it is called a retarding force.

Note
Alternatively, other expressions involving acceleration can also be used to derive its unit. For example, the famous, $s = \dfrac{1}{2}a{t^2}$
Making $a$ subject of the formula, we have,
$\Rightarrow a = \dfrac{{2s}}{{{t^2}}}$, replacing the quantities with their unit we have
$\Rightarrow {a_u} = \dfrac{m}{{{s^2}}} = m{s^{ - 2}}$ (2 has no unit). The result is identical to the one given in the solution section.
It is best to directly use expressions whose variables have SI units contained in the options.