Question

A car running at $54{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ is slowed down by $36{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ in 4 seconds by the application of brakes. Calculate the retardation produced.
A. $9{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{h}}{{\rm{r}}^2}}}} \right. } {{\rm{h}}{{\rm{r}}^2}}}$
B. $10{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{h}}{{\rm{r}}^2}}}} \right. } {{\rm{h}}{{\rm{r}}^2}}}$
C. $12{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{h}}{{\rm{r}}^2}}}} \right. } {{\rm{h}}{{\rm{r}}^2}}}$
D. $18{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{h}}{{\rm{r}}^2}}}} \right. } {{\rm{h}}{{\rm{r}}^2}}}$

Answer 1

We will use the first equation of motion which gives us the relationship between initial velocity, final velocity, time taken by car to reach final velocity and retardation caused by the application of brakes.

Complete step by step answer:
Given:
The initial velocity of car is ${V_1} = 54{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$.
The car is slowed down by a speed is $x = 36{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$.
The time period taken by car to reduce its x velocity is $t = 4{\rm{ s}}$.
We have to find the value of retardation produced due to the application of brakes.
It is given that the car is slowed down by x speed in t seconds of time so we can write the final velocity of car as below:
${V_2} = {V_1} - x$
On substituting $54{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right. } {{\rm{hr}}}}$ for ${V_1}$ and $36{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ for ${V_2}$ in the above expression, we get:

$${V_2} = 54{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}} - 36{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}\\\ = 18{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$$

Let us write the expression for retardation of the car which is given by the first equation of motion.
$v = u - at$
Here, v is the final velocity, u is the initial velocity and a is the retardation of the car.
On rearranging the above expression, we can write:
$a = \dfrac{{u - v}}{t}$
On substituting $54{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ for ${V_1}$, $18{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}$ for ${V_1}$ and $4{\rm{ s}}$ for t in the above expression, we get:

$$\Rightarrow a = \dfrac{{54{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}} - 18{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{hr}}}}} \right.} {{\rm{hr}}}}}}{{4{\rm{ s}} \times \left( {\dfrac{{{\rm{hr}}}}{{3600{\rm{ s}}}}} \right)}}\\\ \Rightarrow a = 32400{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{h}}{{\rm{r}}^2}}}} \right.} {{\rm{h}}{{\rm{r}}^2}}}$$

Therefore, the retardation produced due to application of brakes is 32400 ${\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {{\rm{h}}{{\rm{r}}^2}}}} \right.} {{\rm{h}}{{\rm{r}}^2}}}$ and the given options are incorrect.

Note: Unit conversion is an important step in this problem. We know that there are 60 minutes in one hour and 60 seconds in one minute so we can say that there are 3600 seconds in one hour. Do not forget to convert the second into hour to follow the system of homogeneity of units.