Question

A car travelling with a velocity of $80\,km/h$ slowed down to $44\,km/h$ in $15\,s$.The retardation is
A. $0.67{\text{ m/}}{{\text{s}}^2}$
B. $1{\text{ m/}}{{\text{s}}^2}$
C. $1.25{\text{ m/}}{{\text{s}}^2}$
D. $1.5{\text{ m/}}{{\text{s}}^2}$

Answer 1

Hint:To solve this question students should know the general equations of motion. Using the general equation of motion with the given details from the question we can find the retardation of the car. Retardation is the act or result of delaying; the extent to which anything is retarded or delayed; that which retards or delays.

Complete step by step answer:
From the equations of motion, we have $v = u + at - - - (1)$
Where, $u$ initial velocity of the body, $a$ = uniform acceleration of the body, $v$ = velocity of the body after a time $t$.
So now it’s given that
$u = {\text{ 80km/h = 80 }} \times {\text{ }}\dfrac{5}{{18}}{\text{ m/s = }}\dfrac{{200}}{9}{\text{ m/s}}$
$ \Rightarrow v = {\text{ }}44{\text{ km/h = 44 }} \times {\text{ }}\dfrac{5}{{18}}{\text{ m/s = }}\dfrac{{110}}{9}{\text{ m/s}}$
Time (t) = 15 seconds

So, by substituting the above values in equation (1) we get
$v = u + at$
$ \Rightarrow \dfrac{{110}}{9} = \dfrac{{200}}{9} + 15a$
$ \Rightarrow a = \dfrac{{\dfrac{{110}}{9} - \dfrac{{200}}{9}}}{{15}}$
On further simplifying we get
$ \Rightarrow a = \dfrac{{ - 90}}{{9 \times 15}}$
$ \Rightarrow a = \dfrac{{ - 10}}{{15}}$
$\therefore a = - 0.67{\text{ m/}}{{\text{s}}^2}$
That is the retardation is equal to $0.67{\text{ m/}}{{\text{s}}^2}$.

Therefore, the correct answer is option A.

Additional information:The acceleration is the rate of change of velocity or change in velocity per unit time interval. Velocity is a vector quantity hence a change in its magnitude or in direction or in both, will give the acceleration (or non-uniform motion).Equations of motion are valid only when acceleration remains constant during the motion. Otherwise, we have to use the following equations
$v = \int {a\left( {\text{t}} \right)dt{\text{ and }}s = \int {v\left( {\text{t}} \right)dt} } $

Distance between two points is the length of the path between the two points. It is a scalar and its value cannot be negative. It is measured in metres in S.I. unit. The shortest distance between the initial position to final position is called displacement of the object. It is a vector quantity. SI unit of displacement is meter (m). CGS unit of displacement is centimetre (cm). Its direction is from initial position to final position. Speed is related to distance and it is a scalar while velocity is related to displacement and it is a vector. For a moving body speed cannot have zero or negative values but velocity can have.

Note:We here convert km/h into m/s because as the standard unit of velocity is m/s and students should be careful that all units of the given physical quantity should be in the same units. For our convenience of doing the problem here we have converted km/h to m/s.We can convert km/h to m/s by
$1\left( {{\text{km/h}}} \right) = \dfrac{{{\text{1000}}}}{{{\text{3600}}}}\left( {{\text{m/s}}} \right)$ which can be simplified to $1\left( {{\text{km/h}}} \right) = \dfrac{5}{{18}}\left( {{\text{m/s}}} \right)$