Question

An automobile travelling with speed of 60Km /hr, can brake to stop within a distance of 20m. If the car is going as fast,i.e.120 Km/hr, the stopping distance will be:
A) 20m
B) 40m
C) 60m
D) 80m

Answer 1

Hint: Consider an automobile moving at an initial velocity ‘u’ with acceleration ‘a’. When the vehicle will stop, then its Final velocity will be zero. Then, we can find the retardation of the automobile by using the $3^{rd}$ equation of motion. Then, in the $2^{nd}$ case, retardation remains the same and thus ,we can find the value of stopping distance by using the $3^{rd}$ equation of motion.

Formula used:
$3^{rd}$ Equation of motion: ${{\text{v}}^2} = {u^2} + 2as$
Where: ‘$v$’ is Final Velocity, ‘u’ is Initial Velocity, ‘a’ is acceleration, ‘s’ is distance covered.

Complete step-by-step answer:
Case 1: Given, u=60km/hr, s=20m, v=0 m/sec. Here, we will convert the unit of ‘u’ from km/hr to m/sec.
$u = 60km/hr $
$\Rightarrow u = \dfrac{{60 \times 1000}}{{60 \times 60}}{\text{ }}\left( {As{\text{ 1km = 1000m and 1hr = 60}} \times {\text{60sec}}} \right) $
$\Rightarrow u = \dfrac{{50}}{3}m/\sec $
$\therefore u = 16.67m/\sec $
So, ‘u’=16.67m/sec , ‘s’=20m,’ v’=0 m/sec ………………………… (i)
So, we can find the value of ‘a’ by using the $3^{rd}$ equation of motion.
$3^{rd}$ Equation of motion is ${{\text{v}}^2} = {u^2} + 2as$…………………………… (ii)
So, putting the value in equation (ii) from (i), we have:
$
 \Rightarrow {0^2} = {\left( {16.67} \right)^2} + 2 \times a \times 20 \\
 \Rightarrow 40a = - 277.88 \\
 \Rightarrow a = \dfrac{{ - 277.88}}{{40}} \\
 \therefore a = - 6.94m/{s^2} \\
 $
Here, the sign of ‘a’ is negative because the automobile is retarding and going to stop.
Case 2: ‘u’=120km/hr ,’v’ =0 m/sec
Again, we will convert the unit of ‘u’ in m/sec.
$
u = 120km/hr \\
\Rightarrow u = \dfrac{{120 \times 1000}}{{60 \times 60}}{\text{ }}\left( {As{\text{ 1km = 1000m and 1hr = 60}} \times {\text{60sec}}} \right) \\
\Rightarrow u = \dfrac{{100}}{3}m/\sec \\
\therefore u = 33.33m/\sec \\
$
Here, the value of ‘a’ remains the same because the resisting force on an automobile is independent of velocity.
So, ‘u’ =33.33m/sec, ‘v’ =0 m/sec, ‘a’= $-6.94m/{s^2}$ ……………………………… (iii)
Putting the value of (iii) in 3rd equation of motion, we have:
$
   \Rightarrow {0^2} = {\left( {33.33} \right)^2} + 2 \times \left( { - 6.94} \right) \times s \\
   \Rightarrow 13.88s = 1110.88 \\
   \Rightarrow s = \dfrac{{1110.88}}{{13.88}} \\
   \Rightarrow s = 80m \\
$

Thus, the stopping distance will be 80m.

Note: Equations of motion, in physics, are defined as equations that describe the behaviour of a physical system in terms of its motion as a function of time.
There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a). The following are the three equation of motion:
First Equation of Motion: $v = u + at$
Second Equation of Motion: $s = ut + \dfrac{1}{2}a{t^2}$
Third Equation of Motion: ${{\text{v}}^2} = {u^2} + 2as$